3.13.45 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx\) [1245]

3.13.45.1 Optimal result

Integrand size = 27, antiderivative size = 243 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 f}+\frac {i (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 f}-\frac {(b c-a d)^{3/2} \left (4 a b c+a^2 d+5 b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 f}-\frac {(b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))} \]

output

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

3.13.45.2 Mathematica [A] (verified)

Time = 5.24 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {-\frac {i (a+i b)^2 b^{3/2} (c-i d)^{5/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+i b^{3/2} (i a+b)^2 (c+i d)^{5/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+(b c-a d)^{5/2} \left (4 a b c+a^2 d+5 b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right )}+\frac {d (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b}+d (b c-a d) (c+d \tan (e+f x))^{3/2}+b d (c+d \tan (e+f x))^{5/2}-\frac {b^2 (c+d \tan (e+f x))^{7/2}}{a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d) f} \]

input

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

output

(-((I*(a + I*b)^2*b^(3/2)*(c - I*d)^(5/2)*(b*c - a*d)*ArcTanh[Sqrt[c + d*T 
an[e + f*x]]/Sqrt[c - I*d]] + I*b^(3/2)*(I*a + b)^2*(c + I*d)^(5/2)*(b*c - 
 a*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + (b*c - a*d)^(5/2)* 
(4*a*b*c + a^2*d + 5*b^2*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqr 
t[b*c - a*d]])/(b^(3/2)*(a^2 + b^2))) + (d*(b*c - a*d)^2*Sqrt[c + d*Tan[e 
+ f*x]])/b + d*(b*c - a*d)*(c + d*Tan[e + f*x])^(3/2) + b*d*(c + d*Tan[e + 
 f*x])^(5/2) - (b^2*(c + d*Tan[e + f*x])^(7/2))/(a + b*Tan[e + f*x]))/((a^ 
2 + b^2)*(b*c - a*d)*f)
 

3.13.45.3 Rubi [A] (warning: unable to verify)

Time = 2.02 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 4048, 27, 3042, 4136, 27, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

\(\displaystyle -\frac {(b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {\frac {(b c-a d)^2 \left (a^2 d+4 a b c+5 b^2 d\right ) \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}-\frac {2 \left (\frac {b (-b+i a)^2 (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (b+i a)^2 (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}\right )}{a^2+b^2}}{2 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {(b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {\frac {2 (b c-a d)^2 \left (a^2 d+4 a b c+5 b^2 d\right ) \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}-\frac {2 \left (\frac {b (-b+i a)^2 (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (b+i a)^2 (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}\right )}{a^2+b^2}}{2 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {-\frac {2 \left (a^2 d+4 a b c+5 b^2 d\right ) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} f \left (a^2+b^2\right )}-\frac {2 \left (\frac {b (-b+i a)^2 (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (b+i a)^2 (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}\right )}{a^2+b^2}}{2 b \left (a^2+b^2\right )}\)

input

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

output

((-2*(((I*a - b)^2*b*(c - I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f 
 + (b*(I*a + b)^2*(c + I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f))/ 
(a^2 + b^2) - (2*(b*c - a*d)^(3/2)*(4*a*b*c + a^2*d + 5*b^2*d)*ArcTanh[(Sq 
rt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*f)) 
/(2*b*(a^2 + b^2)) - ((b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(b*(a^2 + b^ 
2)*f*(a + b*Tan[e + f*x]))
 

3.13.45.3.1 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 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 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 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]
 

3.13.45.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4576\) vs. \(2(211)=422\).

Time = 0.90 (sec) , antiderivative size = 4577, normalized size of antiderivative = 18.84

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(4577\)
default	\(\text {Expression too large to display}\)	\(4577\)

input

int((c+d*tan(f*x+e))^(5/2)/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

output

1/4/f/d/(a^2+b^2)^2*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^ 
(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^ 
(1/2)*b^2*c^2-2/f*d/(a^2+b^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c 
+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c 
)^(1/2))*(c^2+d^2)^(1/2)*a^2*c-7/f*d/(a^2+b^2)^2*b/((a*d-b*c)*b)^(1/2)*arc 
tan(b*(c+d*tan(f*x+e))^(1/2)/((a*d-b*c)*b)^(1/2))*a^2*c^2-10/f*d^2/(a^2+b^ 
2)^2*b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(c+d*tan(f*x+e))^(1/2)/((a*d-b*c)*b) 
^(1/2))*a*c+1/4/f/d/(a^2+b^2)^2*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*( 
2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2 
)*(c^2+d^2)^(1/2)*a^2*c^2-1/4/f/d/(a^2+b^2)^2*ln(d*tan(f*x+e)+c-(c+d*tan(f 
*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^( 
1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b^2*c^2+2/f/(a^2+b^2)^2/(2*(c^2+d^2)^(1/2) 
-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2) 
)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a*b*c^2+1/f/(a^2+b^2)^2*l 
n(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2 
+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b*c+2/f/(a^2+ 
b^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*( 
c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)* 
a*b*c^2-1/4/f/d/(a^2+b^2)^2*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c 
^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)...
 

3.13.45.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9974 vs. \(2 (203) = 406\).

Time = 123.10 (sec) , antiderivative size = 19967, normalized size of antiderivative = 82.17 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.13.45.6 Sympy [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]

input

integrate((c+d*tan(f*x+e))**(5/2)/(a+b*tan(f*x+e))**2,x)
 

output

Integral((c + d*tan(e + f*x))**(5/2)/(a + b*tan(e + f*x))**2, x)
 

3.13.45.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]

input

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

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

3.13.45.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.13.45.9 Mupad [B] (verification not implemented)

Time = 19.12 (sec) , antiderivative size = 78132, normalized size of antiderivative = 321.53 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Too large to display} \]

input

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

output

- atan(((((8*(4*b^12*c*d^17*f^2 - 16*a^11*b*d^18*f^2 - 8*a^12*c*d^17*f^2 - 
 4*a*b^11*d^18*f^2 + 304*a^3*b^9*d^18*f^2 + 120*a^5*b^7*d^18*f^2 - 320*a^7 
*b^5*d^18*f^2 - 148*a^9*b^3*d^18*f^2 - 8*a^12*c^3*d^15*f^2 - 400*b^12*c^3* 
d^15*f^2 + 176*b^12*c^5*d^13*f^2 + 488*b^12*c^7*d^11*f^2 - 92*b^12*c^9*d^9 
*f^2 + 6576*a^2*b^10*c^3*d^15*f^2 + 400*a^2*b^10*c^5*d^13*f^2 - 6144*a^2*b 
^10*c^7*d^11*f^2 + 1056*a^2*b^10*c^9*d^9*f^2 - 5408*a^3*b^9*c^2*d^16*f^2 + 
 3888*a^3*b^9*c^4*d^14*f^2 + 6400*a^3*b^9*c^6*d^12*f^2 - 3072*a^3*b^9*c^8* 
d^10*f^2 + 128*a^3*b^9*c^10*d^8*f^2 - 648*a^4*b^8*c^3*d^15*f^2 - 1952*a^4* 
b^8*c^5*d^13*f^2 + 208*a^4*b^8*c^7*d^11*f^2 + 200*a^4*b^8*c^9*d^9*f^2 - 42 
88*a^5*b^7*c^2*d^16*f^2 + 4112*a^5*b^7*c^4*d^14*f^2 + 5120*a^5*b^7*c^6*d^1 
2*f^2 - 3208*a^5*b^7*c^8*d^10*f^2 + 192*a^5*b^7*c^10*d^8*f^2 - 6688*a^6*b^ 
6*c^3*d^15*f^2 - 2464*a^6*b^6*c^5*d^13*f^2 + 5952*a^6*b^6*c^7*d^11*f^2 - 9 
60*a^6*b^6*c^9*d^9*f^2 + 2624*a^7*b^5*c^2*d^16*f^2 - 2016*a^7*b^5*c^4*d^14 
*f^2 - 3456*a^7*b^5*c^6*d^12*f^2 + 1504*a^7*b^5*c^8*d^10*f^2 + 992*a^8*b^4 
*c^3*d^15*f^2 - 144*a^8*b^4*c^5*d^13*f^2 - 888*a^8*b^4*c^7*d^11*f^2 - 12*a 
^8*b^4*c^9*d^9*f^2 + 352*a^9*b^3*c^2*d^16*f^2 + 520*a^9*b^3*c^4*d^14*f^2 + 
 32*a^9*b^3*c^6*d^12*f^2 + 12*a^9*b^3*c^8*d^10*f^2 + 48*a^10*b^2*c^3*d^15* 
f^2 + 144*a^10*b^2*c^5*d^13*f^2 + 1120*a*b^11*c^2*d^16*f^2 - 2776*a*b^11*c 
^4*d^14*f^2 - 2208*a*b^11*c^6*d^12*f^2 + 1628*a*b^11*c^8*d^10*f^2 - 64*a*b 
^11*c^10*d^8*f^2 - 1024*a^2*b^10*c*d^17*f^2 + 1312*a^4*b^8*c*d^17*f^2 +...