3.5.3 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [403]

3.5.3.1 Optimal result

Integrand size = 33, antiderivative size = 325 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {(b (A-B)-a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(b (A-B)-a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 b^{5/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{a^2 d \sqrt {\tan (c+d x)}} \]

output

2*b^(5/2)*(A*b-B*a)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/a^(5/2)/(a^2+ 
b^2)/d+1/2*(b*(A-B)-a*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2) 
/d*2^(1/2)+1/2*(b*(A-B)-a*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b 
^2)/d*2^(1/2)+1/4*(a*(A-B)+b*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+ 
c))/(a^2+b^2)/d*2^(1/2)-1/4*(a*(A-B)+b*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2 
)+tan(d*x+c))/(a^2+b^2)/d*2^(1/2)+2*(A*b-B*a)/a^2/d/tan(d*x+c)^(1/2)-2/3*A 
/a/d/tan(d*x+c)^(3/2)
 

3.5.3.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.63 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.54 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {\frac {3 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{a-i b}+\frac {6 b^{5/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} \left (a^2+b^2\right )}+\frac {3 \sqrt [4]{-1} (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{a+i b}-\frac {2 (a A+(-3 A b+3 a B) \tan (c+d x))}{a^2 \tan ^{\frac {3}{2}}(c+d x)}}{3 d} \]

input

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x 
]
 

output

((3*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/(a - I*b) 
+ (6*b^(5/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a^ 
(5/2)*(a^2 + b^2)) + (3*(-1)^(1/4)*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c 
 + d*x]]])/(a + I*b) - (2*(a*A + (-3*A*b + 3*a*B)*Tan[c + d*x]))/(a^2*Tan[ 
c + d*x]^(3/2)))/(3*d)
 

3.5.3.3 Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.89, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4092, 27, 3042, 4132, 27, 3042, 4136, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{5/2} (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 4092

\(\displaystyle -\frac {2 \int \frac {3 \left (A b \tan ^2(c+d x)+a A \tan (c+d x)+A b-a B\right )}{2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}dx}{3 a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {A b \tan ^2(c+d x)+a A \tan (c+d x)+A b-a B}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}dx}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {A b \tan (c+d x)^2+a A \tan (c+d x)+A b-a B}{\tan (c+d x)^{3/2} (a+b \tan (c+d x))}dx}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {-\frac {2 \int -\frac {A a^2+B \tan (c+d x) a^2+b B a-A b^2-b (A b-a B) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {A a^2+B \tan (c+d x) a^2+b B a-A b^2-b (A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {A a^2+B \tan (c+d x) a^2+b B a-A b^2-b (A b-a B) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4136

\(\displaystyle -\frac {\frac {\frac {\int \frac {a^2 (a A+b B)-a^2 (A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b^3 (A b-a B) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\int \frac {a^2 (a A+b B)-a^2 (A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4017

\(\displaystyle -\frac {\frac {\frac {2 \int \frac {a^2 (a A+b B-(A b-a B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {2 a^2 \int \frac {a A+b B-(A b-a B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1482

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4117

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^3 (A b-a B) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {2 b^3 (A b-a B) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {2 b^{5/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}}{a}-\frac {2 (A b-a B)}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

input

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]
 

output

-((((-2*b^(5/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/ 
(Sqrt[a]*(a^2 + b^2)*d) + (2*a^2*(-1/2*((b*(A - B) - a*(A + B))*(-(ArcTan[ 
1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + 
 d*x]]]/Sqrt[2])) + ((a*(A - B) + b*(A + B))*(-1/2*Log[1 - Sqrt[2]*Sqrt[Ta 
n[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + 
 Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/a - (2*(A*b - a*B))/(a*d 
*Sqrt[Tan[c + d*x]]))/a) - (2*A)/(3*a*d*Tan[c + d*x]^(3/2))
 

3.5.3.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4092

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) 
/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 
2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* 
B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 
)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n 
+ 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] 
 || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ 
[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

3.5.3.4 Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {2 b^{3} \left (A b -B a \right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{a^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}-\frac {2 A}{3 a \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-A b +B a \right )}{a^{2} \sqrt {\tan \left (d x +c \right )}}+\frac {\frac {\left (-a A -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(286\)
default	\(\frac {\frac {2 b^{3} \left (A b -B a \right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{a^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}-\frac {2 A}{3 a \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-A b +B a \right )}{a^{2} \sqrt {\tan \left (d x +c \right )}}+\frac {\frac {\left (-a A -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(286\)

input

int((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

output

1/d*(2/a^2*b^3*(A*b-B*a)/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/( 
a*b)^(1/2))-2/3/a*A/tan(d*x+c)^(3/2)-2/a^2*(-A*b+B*a)/tan(d*x+c)^(1/2)+2/( 
a^2+b^2)*(1/8*(-A*a-B*b)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c 
))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^ 
(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/8*(A*b-B*a)*2^(1/2)*(ln((1 
-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+ 
c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^( 
1/2)))))
 

3.5.3.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3129 vs. \(2 (281) = 562\).

Time = 27.01 (sec) , antiderivative size = 6284, normalized size of antiderivative = 19.34 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm= 
"fricas")
 

output

Too large to include
 

3.5.3.6 Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Timed out} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c)),x)
 

output

Timed out
 

3.5.3.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.79 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {\frac {24 \, {\left (B a b^{3} - A b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a b}} + \frac {3 \, {\left (2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{2} + b^{2}} + \frac {8 \, {\left (A a + 3 \, {\left (B a - A b\right )} \tan \left (d x + c\right )\right )}}{a^{2} \tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm= 
"maxima")
 

output

-1/12*(24*(B*a*b^3 - A*b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^4 + 
 a^2*b^2)*sqrt(a*b)) + 3*(2*sqrt(2)*((A + B)*a - (A - B)*b)*arctan(1/2*sqr 
t(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A + B)*a - (A - B)*b) 
*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*((A - B)* 
a + (A + B)*b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2 
)*((A - B)*a + (A + B)*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 
 1))/(a^2 + b^2) + 8*(A*a + 3*(B*a - A*b)*tan(d*x + c))/(a^2*tan(d*x + c)^ 
(3/2)))/d
 

3.5.3.8 Giac [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Timed out} \]

input

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm= 
"giac")
 

output

Timed out
 

3.5.3.9 Mupad [B] (verification not implemented)

Time = 15.15 (sec) , antiderivative size = 16111, normalized size of antiderivative = 49.57 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]

input

int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))),x)
 

output

atan(((tan(c + d*x)^(1/2)*(64*A^4*a^14*b^9*d^5 + 32*A^4*a^18*b^5*d^5) + (- 
((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/ 
2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(((-(( 
64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) 
 - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(tan(c + 
 d*x)^(1/2)*(-((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2 
*b^2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)) 
)^(1/2)*(512*a^18*b^9*d^9 + 512*a^20*b^7*d^9 - 512*a^22*b^5*d^9 - 512*a^24 
*b^3*d^9) - 512*A*a^16*b^10*d^8 - 512*A*a^18*b^8*d^8 + 384*A*a^20*b^6*d^8 
+ 256*A*a^22*b^4*d^8 - 128*A*a^24*b^2*d^8) - tan(c + d*x)^(1/2)*(512*A^2*a 
^15*b^10*d^7 + 448*A^2*a^19*b^6*d^7 - 128*A^2*a^21*b^4*d^7 - 64*A^2*a^23*b 
^2*d^7))*(-((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^ 
2*d^4))^(1/2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^( 
1/2) + 384*A^3*a^15*b^9*d^6 - 32*A^3*a^19*b^5*d^6 - 32*A^3*a^21*b^3*d^6))* 
(-((64*A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^( 
1/2) - 8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*1i + 
 (tan(c + d*x)^(1/2)*(64*A^4*a^14*b^9*d^5 + 32*A^4*a^18*b^5*d^5) + (-((64* 
A^4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 
8*A^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(((-((64*A^ 
4*a^2*b^2*d^4 - A^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) -...