\(\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\) [653]

Optimal result

Integrand size = 25, antiderivative size = 265 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \] Output:

I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(5 
/2)/d-I*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I* 
a+b)^(5/2)/d-2/a/d/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2)-2/3*b*(3*a^2+4* 
b^2)*tan(d*x+c)^(1/2)/a^2/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)-2/3*b*(3*a^4+ 
17*a^2*b^2+8*b^4)*tan(d*x+c)^(1/2)/a^3/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2 
)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.92 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=-\frac {-\frac {3 \sqrt [4]{-1} a^2 \left (\frac {(a+i b)^2 \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}-\frac {(a-i b)^2 \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{\left (a^2+b^2\right )^2}+\frac {6 a}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}}{3 a^2 d} \] Input:

Integrate[1/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(5/2)),x]
 

Output:

-1/3*((-3*(-1)^(1/4)*a^2*(((a + I*b)^2*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*S 
qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] - ((a - I*b)^ 
2*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
d*x]]])/Sqrt[a + I*b]))/(a^2 + b^2)^2 + (6*a)/(Sqrt[Tan[c + d*x]]*(a + b*T 
an[c + d*x])^(3/2)) + (2*b*(3*a^2 + 4*b^2)*Sqrt[Tan[c + d*x]])/((a^2 + b^2 
)*(a + b*Tan[c + d*x])^(3/2)) + (2*b*(3*a^4 + 17*a^2*b^2 + 8*b^4)*Sqrt[Tan 
[c + d*x]])/(a*(a^2 + b^2)^2*Sqrt[a + b*Tan[c + d*x]]))/(a^2*d)
 

Rubi [A] (verified)

Time = 1.67 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.23, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

\(\displaystyle \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^{3/2} (a+b \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4052

\(\displaystyle -\frac {2 \int \frac {4 b \tan ^2(c+d x)+a \tan (c+d x)+4 b}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}dx}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {4 b \tan ^2(c+d x)+a \tan (c+d x)+4 b}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}dx}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {4 b \tan (c+d x)^2+a \tan (c+d x)+4 b}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}dx}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {\frac {2 \int \frac {3 \tan (c+d x) a^3+2 b \left (3 a^2+4 b^2\right ) \tan ^2(c+d x)+b \left (9 a^2+8 b^2\right )}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {3 \tan (c+d x) a^3+2 b \left (3 a^2+4 b^2\right ) \tan ^2(c+d x)+b \left (9 a^2+8 b^2\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {3 \tan (c+d x) a^3+2 b \left (3 a^2+4 b^2\right ) \tan (c+d x)^2+b \left (9 a^2+8 b^2\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {\frac {\frac {2 \int \frac {3 \left (2 b a^4+\left (a^2-b^2\right ) \tan (c+d x) a^3\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {3 \int \frac {2 b a^4+\left (a^2-b^2\right ) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {3 \int \frac {2 b a^4+\left (a^2-b^2\right ) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 a \left (a^2+b^2\right )}+\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a}-\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4099

\(\displaystyle -\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} i a^3 (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i a^3 (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} i a^3 (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i a^3 (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 4098

\(\displaystyle -\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {i a^3 (a-i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {i a^3 (a+i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {i a^3 (a-i b)^2 \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {i a^3 (a+i b)^2 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {i a^3 (a-i b)^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {i a^3 (a+i b)^2 \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2}{a d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\frac {2 b \left (3 a^2+4 b^2\right ) \sqrt {\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {2 b \left (3 a^4+17 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {i a^3 (a-i b)^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {i a^3 (a+i b)^2 \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a \left (a^2+b^2\right )}}{3 a \left (a^2+b^2\right )}}{a}\)

Input:

Int[1/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(5/2)),x]
 

Output:

-2/(a*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) - ((2*b*(3*a^2 + 4* 
b^2)*Sqrt[Tan[c + d*x]])/(3*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + 
((3*((I*a^3*(a - I*b)^2*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
 b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - (I*a^3*(a + I*b)^2*ArcTanh[(Sqrt[I* 
a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/ 
(a*(a^2 + b^2)) + (2*b*(3*a^4 + 17*a^2*b^2 + 8*b^4)*Sqrt[Tan[c + d*x]])/(a 
*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(3*a*(a^2 + b^2)))/a
 

Defintions of rubi rules used

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216

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

rule 219

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

rule 3042

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

rule 4052

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

rule 4098

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[A^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* 
x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099

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[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

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])))
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.19 (sec) , antiderivative size = 1489246, normalized size of antiderivative = 5619.80

\[\text {output too large to display}\]

Input:

int(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11517 vs. \(2 (221) = 442\).

Time = 2.24 (sec) , antiderivative size = 11517, normalized size of antiderivative = 43.46 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate(1/tan(d*x+c)**(3/2)/(a+b*tan(d*x+c))**(5/2),x)
 

Output:

Integral(1/((a + b*tan(c + d*x))**(5/2)*tan(c + d*x)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(3/2)), x)
 

Giac [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate(1/((b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(5/2)),x)
 

Output:

int(1/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx=\frac {-16 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2} b^{2}-24 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right ) a b -6 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, a^{2}-3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) \tan \left (d x +c \right )^{3} a^{3} b^{2} d -6 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) \tan \left (d x +c \right )^{2} a^{4} b d -3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) \tan \left (d x +c \right ) a^{5} d}{3 \tan \left (d x +c \right ) a^{3} d \left (\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}\right )} \] Input:

int(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x)
 

Output:

( - 16*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2*b**2 - 
24*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)*a*b - 6*sqrt(t 
an(c + d*x))*sqrt(tan(c + d*x)*b + a)*a**2 - 3*int((sqrt(tan(c + d*x))*sqr 
t(tan(c + d*x)*b + a))/(tan(c + d*x)**3*b**3 + 3*tan(c + d*x)**2*a*b**2 + 
3*tan(c + d*x)*a**2*b + a**3),x)*tan(c + d*x)**3*a**3*b**2*d - 6*int((sqrt 
(tan(c + d*x))*sqrt(tan(c + d*x)*b + a))/(tan(c + d*x)**3*b**3 + 3*tan(c + 
 d*x)**2*a*b**2 + 3*tan(c + d*x)*a**2*b + a**3),x)*tan(c + d*x)**2*a**4*b* 
d - 3*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a))/(tan(c + d*x)**3*b 
**3 + 3*tan(c + d*x)**2*a*b**2 + 3*tan(c + d*x)*a**2*b + a**3),x)*tan(c + 
d*x)*a**5*d)/(3*tan(c + d*x)*a**3*d*(tan(c + d*x)**2*b**2 + 2*tan(c + d*x) 
*a*b + a**2))