\(\int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1255]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-1)]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 317 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i (a-i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {i (a+i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f} \] Output:

-I*(a-I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f 
+I*(a+I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(3/2)/f 
-2*(-a*d+b*c)^2*(a+b*tan(f*x+e))^2/d/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)-2/ 
3*b*(15*a^2*b*c*d^2-6*a^3*d^3-12*a*b^2*d*(2*c^2+d^2)+b^3*(8*c^3+5*c*d^2))* 
(c+d*tan(f*x+e))^(1/2)/d^3/(c^2+d^2)/f-2/3*b^2*(3*a*d*(-a*d+2*b*c)-b^2*(4* 
c^2+d^2))*tan(f*x+e)*(c+d*tan(f*x+e))^(1/2)/d^2/(c^2+d^2)/f
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.70 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {12 a (a-b) b (a+b) \left (-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )+\frac {2 b^2 \left (28 a b c d-29 a^2 d^2+b^2 \left (-8 c^2+3 d^2\right )\right )}{d^2 \sqrt {c+d \tan (e+f x)}}-\frac {3 \left (-4 a^3 b c+4 a b^3 c+a^4 d-6 a^2 b^2 d+b^4 d\right ) \left ((-i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )+(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right )}{\left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {4 b^2 (2 b c-5 a d) (a+b \tan (e+f x))}{d \sqrt {c+d \tan (e+f x)}}+\frac {2 b^2 (a+b \tan (e+f x))^2}{\sqrt {c+d \tan (e+f x)}}}{3 d f} \] Input:

Integrate[(a + b*Tan[e + f*x])^4/(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

(12*a*(a - b)*b*(a + b)*(((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I 
*d]])/Sqrt[c - I*d] + (I*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/ 
Sqrt[c + I*d]) + (2*b^2*(28*a*b*c*d - 29*a^2*d^2 + b^2*(-8*c^2 + 3*d^2)))/ 
(d^2*Sqrt[c + d*Tan[e + f*x]]) - (3*(-4*a^3*b*c + 4*a*b^3*c + a^4*d - 6*a^ 
2*b^2*d + b^4*d)*(((-I)*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[ 
e + f*x])/(c - I*d)] + (I*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Ta 
n[e + f*x])/(c + I*d)]))/((c^2 + d^2)*Sqrt[c + d*Tan[e + f*x]]) - (4*b^2*( 
2*b*c - 5*a*d)*(a + b*Tan[e + f*x]))/(d*Sqrt[c + d*Tan[e + f*x]]) + (2*b^2 
*(a + b*Tan[e + f*x])^2)/Sqrt[c + d*Tan[e + f*x]])/(3*d*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.99 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4048, 27, 3042, 4120, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {2 \int \frac {(a+b \tan (e+f x)) \left (c d a^3+6 b d^2 a^2-9 b^2 c d a+4 b^3 c^2-b \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan ^2(e+f x)+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(a+b \tan (e+f x)) \left (c d a^3+6 b d^2 a^2-9 b^2 c d a+4 b^3 c^2-b \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan ^2(e+f x)+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(a+b \tan (e+f x)) \left (c d a^3+6 b d^2 a^2-9 b^2 c d a+4 b^3 c^2-b \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x)^2+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {-\frac {2 \int -\frac {3 c d^2 a^4+18 b d^3 a^3-33 b^2 c d^2 a^2+24 b^3 c^2 d a-b \left (\left (8 c^3+5 d^2 c\right ) b^3-12 a d \left (2 c^2+d^2\right ) b^2+15 a^2 c d^2 b-6 a^3 d^3\right ) \tan ^2(e+f x)-2 b^4 c \left (4 c^2+d^2\right )+3 d^2 \left (-d a^4+4 b c a^3+6 b^2 d a^2-4 b^3 c a-b^4 d\right ) \tan (e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {3 c d^2 a^4+18 b d^3 a^3-33 b^2 c d^2 a^2+24 b^3 c^2 d a-b \left (\left (8 c^3+5 d^2 c\right ) b^3-12 a d \left (2 c^2+d^2\right ) b^2+15 a^2 c d^2 b-6 a^3 d^3\right ) \tan ^2(e+f x)-2 b^4 c \left (4 c^2+d^2\right )+3 d^2 \left (-d a^4+4 b c a^3+6 b^2 d a^2-4 b^3 c a-b^4 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {3 c d^2 a^4+18 b d^3 a^3-33 b^2 c d^2 a^2+24 b^3 c^2 d a-b \left (\left (8 c^3+5 d^2 c\right ) b^3-12 a d \left (2 c^2+d^2\right ) b^2+15 a^2 c d^2 b-6 a^3 d^3\right ) \tan (e+f x)^2-2 b^4 c \left (4 c^2+d^2\right )+3 d^2 \left (-d a^4+4 b c a^3+6 b^2 d a^2-4 b^3 c a-b^4 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {\frac {\int \frac {3 \left (c a^4+4 b d a^3-6 b^2 c a^2-4 b^3 d a+b^4 c\right ) d^2+3 \left (-d a^4+4 b c a^3+6 b^2 d a^2-4 b^3 c a-b^4 d\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {3 \left (c a^4+4 b d a^3-6 b^2 c a^2-4 b^3 d a+b^4 c\right ) d^2+3 \left (-d a^4+4 b c a^3+6 b^2 d a^2-4 b^3 c a-b^4 d\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3}{2} d^2 (a-i b)^4 (c+i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {3}{2} d^2 (a+i b)^4 (c-i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3}{2} d^2 (a-i b)^4 (c+i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {3}{2} d^2 (a+i b)^4 (c-i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3 i d^2 (a-i b)^4 (c+i d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {3 i d^2 (a+i b)^4 (c-i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {3 i d^2 (a-i b)^4 (c+i d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {3 i d^2 (a+i b)^4 (c-i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {3 d (a-i b)^4 (c+i d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {3 d (a+i b)^4 (c-i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}+\frac {3 d^2 (a-i b)^4 (c+i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {3 d^2 (a+i b)^4 (c-i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{3 d}}{d \left (c^2+d^2\right )}\)

Input:

Int[(a + b*Tan[e + f*x])^4/(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

(-2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^2)/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[ 
e + f*x]]) + ((-2*b^2*(3*a*d*(2*b*c - a*d) - b^2*(4*c^2 + d^2))*Tan[e + f* 
x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + ((3*(a - I*b)^4*(c + I*d)*d^2*ArcTa 
n[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (3*(a + I*b)^4*(c - I*d 
)*d^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*b*(15*a^2 
*b*c*d^2 - 6*a^3*d^3 - 12*a*b^2*d*(2*c^2 + d^2) + b^3*(8*c^3 + 5*c*d^2))*S 
qrt[c + d*Tan[e + f*x]])/(d*f))/(3*d))/(d*(c^2 + d^2))
 

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 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4020
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4048
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4113
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

rule 4120
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 
 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2))   Int[(c + d*Tan[e + f*x])^n*Si 
mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* 
d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[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[c^2 + d^2, 0] && 
  !LtQ[n, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(9379\) vs. \(2(287)=574\).

Time = 0.58 (sec) , antiderivative size = 9380, normalized size of antiderivative = 29.59

method result size
parts \(\text {Expression too large to display}\) \(9380\)
derivativedivides \(\text {Expression too large to display}\) \(20018\)
default \(\text {Expression too large to display}\) \(20018\)

Input:

int((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11347 vs. \(2 (278) = 556\).

Time = 14.48 (sec) , antiderivative size = 11347, normalized size of antiderivative = 35.79 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{4}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**4/(c+d*tan(f*x+e))**(3/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**4/(c + d*tan(e + f*x))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{4}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)^4/(d*tan(f*x + e) + c)^(3/2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 20.82 (sec) , antiderivative size = 26741, normalized size of antiderivative = 84.36 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(e + f*x))^4/(c + d*tan(e + f*x))^(3/2),x)
 

Output:

(2*b^4*(c + d*tan(e + f*x))^(3/2))/(3*d^3*f) - atan(-(((c + d*tan(e + f*x) 
)^(1/2)*(16*a^8*d^10*f^3 + 16*b^8*d^10*f^3 - 448*a^2*b^6*d^10*f^3 + 1120*a 
^4*b^4*d^10*f^3 - 448*a^6*b^2*d^10*f^3 + 32*a^8*c^2*d^8*f^3 - 32*a^8*c^6*d 
^4*f^3 - 16*a^8*c^8*d^2*f^3 + 32*b^8*c^2*d^8*f^3 - 32*b^8*c^6*d^4*f^3 - 16 
*b^8*c^8*d^2*f^3 - 896*a^2*b^6*c^2*d^8*f^3 + 896*a^2*b^6*c^6*d^4*f^3 + 448 
*a^2*b^6*c^8*d^2*f^3 - 5376*a^3*b^5*c^3*d^7*f^3 - 5376*a^3*b^5*c^5*d^5*f^3 
 - 1792*a^3*b^5*c^7*d^3*f^3 + 2240*a^4*b^4*c^2*d^8*f^3 - 2240*a^4*b^4*c^6* 
d^4*f^3 - 1120*a^4*b^4*c^8*d^2*f^3 + 5376*a^5*b^3*c^3*d^7*f^3 + 5376*a^5*b 
^3*c^5*d^5*f^3 + 1792*a^5*b^3*c^7*d^3*f^3 - 896*a^6*b^2*c^2*d^8*f^3 + 896* 
a^6*b^2*c^6*d^4*f^3 + 448*a^6*b^2*c^8*d^2*f^3 + 256*a*b^7*c*d^9*f^3 - 256* 
a^7*b*c*d^9*f^3 + 768*a*b^7*c^3*d^7*f^3 + 768*a*b^7*c^5*d^5*f^3 + 256*a*b^ 
7*c^7*d^3*f^3 - 1792*a^3*b^5*c*d^9*f^3 + 1792*a^5*b^3*c*d^9*f^3 - 768*a^7* 
b*c^3*d^7*f^3 - 768*a^7*b*c^5*d^5*f^3 - 256*a^7*b*c^7*d^3*f^3) + (-(((8*a^ 
8*c^3*f^2 + 8*b^8*c^3*f^2 + 64*a*b^7*d^3*f^2 - 64*a^7*b*d^3*f^2 - 24*a^8*c 
*d^2*f^2 - 24*b^8*c*d^2*f^2 - 224*a^2*b^6*c^3*f^2 + 560*a^4*b^4*c^3*f^2 - 
224*a^6*b^2*c^3*f^2 - 448*a^3*b^5*d^3*f^2 + 448*a^5*b^3*d^3*f^2 - 192*a*b^ 
7*c^2*d*f^2 + 192*a^7*b*c^2*d*f^2 + 672*a^2*b^6*c*d^2*f^2 + 1344*a^3*b^5*c 
^2*d*f^2 - 1680*a^4*b^4*c*d^2*f^2 - 1344*a^5*b^3*c^2*d*f^2 + 672*a^6*b^2*c 
*d^2*f^2)^2/4 - (16*c^6*f^4 + 16*d^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4 
)*(a^16 + b^16 + 8*a^2*b^14 + 28*a^4*b^12 + 56*a^6*b^10 + 70*a^8*b^8 + ...
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

int((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x)
 

Output:

( - 2*sqrt(tan(e + f*x)*d + c)*a**4 + int((sqrt(tan(e + f*x)*d + c)*tan(e 
+ f*x)**4)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f 
*x)*b**4*d**2*f + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**4)/(tan(e + 
f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*b**4*c*d*f + 4*int((sqrt(tan( 
e + f*x)*d + c)*tan(e + f*x)**3)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c* 
d + c**2),x)*tan(e + f*x)*a*b**3*d**2*f + 4*int((sqrt(tan(e + f*x)*d + c)* 
tan(e + f*x)**3)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*a*b 
**3*c*d*f - int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)** 
2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f*x)*a**4*d**2*f + 6*int((s 
qrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**2*d**2 + 2*tan(e + 
 f*x)*c*d + c**2),x)*tan(e + f*x)*a**2*b**2*d**2*f - int((sqrt(tan(e + f*x 
)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c** 
2),x)*a**4*c*d*f + 6*int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e 
 + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*a**2*b**2*c*d*f + 4*int((s 
qrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**2*d**2 + 2*tan(e + f* 
x)*c*d + c**2),x)*tan(e + f*x)*a**3*b*d**2*f + 4*int((sqrt(tan(e + f*x)*d 
+ c)*tan(e + f*x))/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*a 
**3*b*c*d*f)/(d*f*(tan(e + f*x)*d + c))