3.13.39 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^2} \, dx\) [1239]

3.13.39.1 Optimal result

Integrand size = 27, antiderivative size = 239 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^2} \, dx=-\frac {i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 f}+\frac {i (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 f}-\frac {\sqrt {b c-a d} \left (4 a b c-a^2 d+3 b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \left (a^2+b^2\right )^2 f}-\frac {(b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))} \]

output

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

3.13.39.2 Mathematica [A] (verified)

Time = 3.78 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.32 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {-\frac {4 \left (\frac {3}{4} i (a+i b)^2 b^2 (c-i d)^{3/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )-\frac {3}{4} i (a-i b)^2 b^2 (c+i d)^{3/2} (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\frac {3}{4} b^{3/2} (b c-a d)^{3/2} \left (4 a b c-a^2 d+3 b^2 d\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )\right )}{b^2 \left (a^2+b^2\right )}+3 d (b c-a d) \sqrt {c+d \tan (e+f x)}+3 b d (c+d \tan (e+f x))^{3/2}-\frac {3 b^2 (c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)}}{3 \left (a^2+b^2\right ) (b c-a d) f} \]

input

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

output

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

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

Time = 1.76 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 4050, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4050

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.13.39.3.1 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 4050

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(a + b*Tan[e + f*x])^(m + 
 1)*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m 
 + 1)*(a^2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^ 
(n - 2)*Simp[a*c^2*(m + 1) + a*d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2 
*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*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] && LtQ[m, -1] && LtQ[1, n, 2] && 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]
 

3.13.39.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3205\) vs. \(2(207)=414\).

Time = 0.84 (sec) , antiderivative size = 3206, normalized size of antiderivative = 13.41

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3206\)
default	\(\text {Expression too large to display}\)	\(3206\)

input

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

3.13.39.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6914 vs. \(2 (199) = 398\).

Time = 13.26 (sec) , antiderivative size = 13847, normalized size of antiderivative = 57.94 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.13.39.6 Sympy [F]

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

input

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

output

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

3.13.39.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]

input

integrate((c+d*tan(f*x+e))^(3/2)/(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
 

3.13.39.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.13.39.9 Mupad [B] (verification not implemented)

Time = 14.95 (sec) , antiderivative size = 39388, normalized size of antiderivative = 164.80 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Too large to display} \]

input

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

output

(atan(((((16*(c + d*tan(e + f*x))^(1/2)*(2*b^9*d^16 + a^8*b*d^16 - 5*a^2*b 
^7*d^16 + 17*a^4*b^5*d^16 - 7*a^6*b^3*d^16 - b^9*c^2*d^14 + 66*b^9*c^4*d^1 
2 - b^9*c^6*d^10 + 2*b^9*c^8*d^8 - 204*a*b^8*c^3*d^13 + 234*a*b^8*c^5*d^11 
 - 24*a*b^8*c^7*d^9 - 126*a^3*b^6*c*d^15 + 94*a^5*b^4*c*d^15 - 18*a^7*b^2* 
c*d^15 - 6*a^8*b*c^2*d^14 + a^8*b*c^4*d^12 + 277*a^2*b^7*c^2*d^14 - 715*a^ 
2*b^7*c^4*d^12 + 367*a^2*b^7*c^6*d^10 - 12*a^2*b^7*c^8*d^8 + 892*a^3*b^6*c 
^3*d^13 - 998*a^3*b^6*c^5*d^11 + 192*a^3*b^6*c^7*d^9 - 457*a^4*b^5*c^2*d^1 
4 + 1173*a^4*b^5*c^4*d^12 - 495*a^4*b^5*c^6*d^10 + 18*a^4*b^5*c^8*d^8 - 62 
8*a^5*b^4*c^3*d^13 + 550*a^5*b^4*c^5*d^11 - 40*a^5*b^4*c^7*d^9 + 155*a^6*b 
^3*c^2*d^14 - 285*a^6*b^3*c^4*d^12 + 33*a^6*b^3*c^6*d^10 + 68*a^7*b^2*c^3* 
d^13 - 10*a^7*b^2*c^5*d^11 + 18*a*b^8*c*d^15))/(a^8*f^4 + b^8*f^4 + 4*a^2* 
b^6*f^4 + 6*a^4*b^4*f^4 + 4*a^6*b^2*f^4) + (((16*(78*a^2*b^9*d^15*f^2 - 2* 
a^10*b*d^15*f^2 - 8*a^4*b^7*d^15*f^2 - 60*a^6*b^5*d^15*f^2 + 24*a^8*b^3*d^ 
15*f^2 + 50*b^11*c^2*d^13*f^2 + 22*b^11*c^4*d^11*f^2 - 28*b^11*c^6*d^9*f^2 
 - 546*a^2*b^9*c^2*d^13*f^2 - 296*a^2*b^9*c^4*d^11*f^2 + 328*a^2*b^9*c^6*d 
^9*f^2 - 240*a^3*b^8*c^3*d^12*f^2 - 544*a^3*b^8*c^5*d^10*f^2 + 64*a^3*b^8* 
c^7*d^8*f^2 + 108*a^4*b^7*c^2*d^13*f^2 + 148*a^4*b^7*c^4*d^11*f^2 + 32*a^4 
*b^7*c^6*d^9*f^2 - 296*a^5*b^6*c^3*d^12*f^2 - 456*a^5*b^6*c^5*d^10*f^2 + 9 
6*a^5*b^6*c^7*d^8*f^2 + 580*a^6*b^5*c^2*d^13*f^2 + 312*a^6*b^5*c^4*d^11*f^ 
2 - 328*a^6*b^5*c^6*d^9*f^2 + 144*a^7*b^4*c^3*d^12*f^2 + 352*a^7*b^4*c^...