\(\int \frac {1}{(b \tan ^3(e+f x))^{5/2}} \, dx\) [12]

Optimal result

Integrand size = 14, antiderivative size = 299 \[ \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (e+f x)}}{1+\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}} \] Output:

-2/5*cot(f*x+e)/b^2/f/(b*tan(f*x+e)^3)^(1/2)+2/9*cot(f*x+e)^3/b^2/f/(b*tan 
(f*x+e)^3)^(1/2)-2/13*cot(f*x+e)^5/b^2/f/(b*tan(f*x+e)^3)^(1/2)+2*tan(f*x+ 
e)/b^2/f/(b*tan(f*x+e)^3)^(1/2)+1/2*arctan(-1+2^(1/2)*tan(f*x+e)^(1/2))*ta 
n(f*x+e)^(3/2)*2^(1/2)/b^2/f/(b*tan(f*x+e)^3)^(1/2)+1/2*arctan(1+2^(1/2)*t 
an(f*x+e)^(1/2))*tan(f*x+e)^(3/2)*2^(1/2)/b^2/f/(b*tan(f*x+e)^3)^(1/2)-1/2 
*arctanh(2^(1/2)*tan(f*x+e)^(1/2)/(1+tan(f*x+e)))*tan(f*x+e)^(3/2)*2^(1/2) 
/b^2/f/(b*tan(f*x+e)^3)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx=\frac {-234 \cot (e+f x)+130 \cot ^3(e+f x)-90 \cot ^5(e+f x)+585 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) (-\tan (e+f x))^{5/4} \sqrt [4]{\tan (e+f x)}+1170 \tan (e+f x)+585 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \sqrt [4]{-\tan (e+f x)} \tan ^{\frac {5}{4}}(e+f x)}{585 b^2 f \sqrt {b \tan ^3(e+f x)}} \] Input:

Integrate[(b*Tan[e + f*x]^3)^(-5/2),x]
 

Output:

(-234*Cot[e + f*x] + 130*Cot[e + f*x]^3 - 90*Cot[e + f*x]^5 + 585*ArcTanh[ 
(-Tan[e + f*x]^2)^(1/4)]*(-Tan[e + f*x])^(5/4)*Tan[e + f*x]^(1/4) + 1170*T 
an[e + f*x] + 585*ArcTan[(-Tan[e + f*x]^2)^(1/4)]*(-Tan[e + f*x])^(1/4)*Ta 
n[e + f*x]^(5/4))/(585*b^2*f*Sqrt[b*Tan[e + f*x]^3])
 

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.78, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.500, Rules used = {3042, 4141, 3042, 3955, 3042, 3955, 3042, 3955, 3042, 3955, 3042, 3957, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (b \tan (e+f x)^3\right )^{5/2}}dx\)

\(\Big \downarrow \) 4141

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \int \frac {1}{\tan ^{\frac {15}{2}}(e+f x)}dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \int \frac {1}{\tan (e+f x)^{15/2}}dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (-\int \frac {1}{\tan ^{\frac {11}{2}}(e+f x)}dx-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (-\int \frac {1}{\tan (e+f x)^{11/2}}dx-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\int \frac {1}{\tan ^{\frac {7}{2}}(e+f x)}dx+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\int \frac {1}{\tan (e+f x)^{7/2}}dx+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (-\int \frac {1}{\tan ^{\frac {3}{2}}(e+f x)}dx-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (-\int \frac {1}{\tan (e+f x)^{3/2}}dx-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3955

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\int \sqrt {\tan (e+f x)}dx-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\int \sqrt {\tan (e+f x)}dx-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {\int \frac {\sqrt {\tan (e+f x)}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \int \frac {\tan (e+f x)}{\tan ^2(e+f x)+1}d\sqrt {\tan (e+f x)}}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \int \frac {\tan (e+f x)+1}{\tan ^2(e+f x)+1}d\sqrt {\tan (e+f x)}-\frac {1}{2} \int \frac {1-\tan (e+f x)}{\tan ^2(e+f x)+1}d\sqrt {\tan (e+f x)}\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}\right )-\frac {1}{2} \int \frac {1-\tan (e+f x)}{\tan ^2(e+f x)+1}d\sqrt {\tan (e+f x)}\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\tan (e+f x)-1}d\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (e+f x)-1}d\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\tan (e+f x)}{\tan ^2(e+f x)+1}d\sqrt {\tan (e+f x)}\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\tan (e+f x)}{\tan ^2(e+f x)+1}d\sqrt {\tan (e+f x)}\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (e+f x)}}{\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2}}\right )\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (e+f x)}}{\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2}}\right )\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (e+f x)}}{\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (e+f x)}+1}{\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1}d\sqrt {\tan (e+f x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2}}\right )\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\tan ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{2 \sqrt {2}}\right )\right )}{f}-\frac {2}{5 f \tan ^{\frac {5}{2}}(e+f x)}+\frac {2}{9 f \tan ^{\frac {9}{2}}(e+f x)}-\frac {2}{13 f \tan ^{\frac {13}{2}}(e+f x)}+\frac {2}{f \sqrt {\tan (e+f x)}}\right )}{b^2 \sqrt {b \tan ^3(e+f x)}}\)

Input:

Int[(b*Tan[e + f*x]^3)^(-5/2),x]
 

Output:

(((2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt 
[2]*Sqrt[Tan[e + f*x]]]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + 
 Tan[e + f*x]]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + 
f*x]]/(2*Sqrt[2]))/2))/f - 2/(13*f*Tan[e + f*x]^(13/2)) + 2/(9*f*Tan[e + f 
*x]^(9/2)) - 2/(5*f*Tan[e + f*x]^(5/2)) + 2/(f*Sqrt[Tan[e + f*x]]))*Tan[e 
+ f*x]^(3/2))/(b^2*Sqrt[b*Tan[e + f*x]^3])
 

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 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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, 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 3042

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

rule 3955

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x] 
)^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2   Int[(b*Tan[c + d*x])^(n + 2), x] 
, x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4141

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\tan \left (f x +e \right ) \left (585 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}-b \tan \left (f x +e \right )-\sqrt {b^{2}}}{b \tan \left (f x +e \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+1170 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+1170 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+4680 b^{6} \tan \left (f x +e \right )^{6} \left (b^{2}\right )^{\frac {1}{4}}-936 b^{6} \tan \left (f x +e \right )^{4} \left (b^{2}\right )^{\frac {1}{4}}+520 b^{6} \tan \left (f x +e \right )^{2} \left (b^{2}\right )^{\frac {1}{4}}-360 b^{6} \left (b^{2}\right )^{\frac {1}{4}}\right )}{2340 f \,b^{6} \left (b \tan \left (f x +e \right )^{3}\right )^{\frac {5}{2}} \left (b^{2}\right )^{\frac {1}{4}}}\)	\(272\)
default	\(\frac {\tan \left (f x +e \right ) \left (585 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}-b \tan \left (f x +e \right )-\sqrt {b^{2}}}{b \tan \left (f x +e \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+1170 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+1170 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+4680 b^{6} \tan \left (f x +e \right )^{6} \left (b^{2}\right )^{\frac {1}{4}}-936 b^{6} \tan \left (f x +e \right )^{4} \left (b^{2}\right )^{\frac {1}{4}}+520 b^{6} \tan \left (f x +e \right )^{2} \left (b^{2}\right )^{\frac {1}{4}}-360 b^{6} \left (b^{2}\right )^{\frac {1}{4}}\right )}{2340 f \,b^{6} \left (b \tan \left (f x +e \right )^{3}\right )^{\frac {5}{2}} \left (b^{2}\right )^{\frac {1}{4}}}\)	\(272\)

Input:

int(1/(b*tan(f*x+e)^3)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/2340/f*tan(f*x+e)/b^6*(585*2^(1/2)*(b*tan(f*x+e))^(13/2)*ln(-((b^2)^(1/4 
)*(b*tan(f*x+e))^(1/2)*2^(1/2)-b*tan(f*x+e)-(b^2)^(1/2))/(b*tan(f*x+e)+(b^ 
2)^(1/4)*(b*tan(f*x+e))^(1/2)*2^(1/2)+(b^2)^(1/2)))+1170*2^(1/2)*(b*tan(f* 
x+e))^(13/2)*arctan((2^(1/2)*(b*tan(f*x+e))^(1/2)+(b^2)^(1/4))/(b^2)^(1/4) 
)+1170*2^(1/2)*(b*tan(f*x+e))^(13/2)*arctan((2^(1/2)*(b*tan(f*x+e))^(1/2)- 
(b^2)^(1/4))/(b^2)^(1/4))+4680*b^6*tan(f*x+e)^6*(b^2)^(1/4)-936*b^6*tan(f* 
x+e)^4*(b^2)^(1/4)+520*b^6*tan(f*x+e)^2*(b^2)^(1/4)-360*b^6*(b^2)^(1/4))/( 
b*tan(f*x+e)^3)^(5/2)/(b^2)^(1/4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx=\frac {1170 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\frac {\sqrt {2} \sqrt {b \tan \left (f x + e\right )^{3}}}{\sqrt {b}} + \tan \left (f x + e\right )}{\tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{8} + 1170 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\frac {\sqrt {2} \sqrt {b \tan \left (f x + e\right )^{3}}}{\sqrt {b}} - \tan \left (f x + e\right )}{\tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{8} - 585 \, \sqrt {2} \sqrt {b} \log \left (\frac {\tan \left (f x + e\right )^{2} + \frac {\sqrt {2} \sqrt {b \tan \left (f x + e\right )^{3}}}{\sqrt {b}} + \tan \left (f x + e\right )}{\tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{8} + 585 \, \sqrt {2} \sqrt {b} \log \left (\frac {\tan \left (f x + e\right )^{2} - \frac {\sqrt {2} \sqrt {b \tan \left (f x + e\right )^{3}}}{\sqrt {b}} + \tan \left (f x + e\right )}{\tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{8} + 8 \, {\left (585 \, \tan \left (f x + e\right )^{6} - 117 \, \tan \left (f x + e\right )^{4} + 65 \, \tan \left (f x + e\right )^{2} - 45\right )} \sqrt {b \tan \left (f x + e\right )^{3}}}{2340 \, b^{3} f \tan \left (f x + e\right )^{8}} \] Input:

integrate(1/(b*tan(f*x+e)^3)^(5/2),x, algorithm="fricas")
 

Output:

1/2340*(1170*sqrt(2)*sqrt(b)*arctan((sqrt(2)*sqrt(b*tan(f*x + e)^3)/sqrt(b 
) + tan(f*x + e))/tan(f*x + e))*tan(f*x + e)^8 + 1170*sqrt(2)*sqrt(b)*arct 
an((sqrt(2)*sqrt(b*tan(f*x + e)^3)/sqrt(b) - tan(f*x + e))/tan(f*x + e))*t 
an(f*x + e)^8 - 585*sqrt(2)*sqrt(b)*log((tan(f*x + e)^2 + sqrt(2)*sqrt(b*t 
an(f*x + e)^3)/sqrt(b) + tan(f*x + e))/tan(f*x + e))*tan(f*x + e)^8 + 585* 
sqrt(2)*sqrt(b)*log((tan(f*x + e)^2 - sqrt(2)*sqrt(b*tan(f*x + e)^3)/sqrt( 
b) + tan(f*x + e))/tan(f*x + e))*tan(f*x + e)^8 + 8*(585*tan(f*x + e)^6 - 
117*tan(f*x + e)^4 + 65*tan(f*x + e)^2 - 45)*sqrt(b*tan(f*x + e)^3))/(b^3* 
f*tan(f*x + e)^8)
 

Sympy [F]

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

integrate(1/(b*tan(f*x+e)**3)**(5/2),x)
 

Output:

Integral((b*tan(e + f*x)**3)**(-5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx=\frac {\frac {585 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (f x + e\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (f x + e\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right )\right )}}{b^{\frac {5}{2}}} + \frac {8 \, {\left (\frac {585 \, \sqrt {b}}{\sqrt {\tan \left (f x + e\right )}} - \frac {117 \, \sqrt {b}}{\tan \left (f x + e\right )^{\frac {5}{2}}} + \frac {65 \, \sqrt {b}}{\tan \left (f x + e\right )^{\frac {9}{2}}} - \frac {45 \, \sqrt {b}}{\tan \left (f x + e\right )^{\frac {13}{2}}}\right )}}{b^{3}}}{2340 \, f} \] Input:

integrate(1/(b*tan(f*x+e)^3)^(5/2),x, algorithm="maxima")
 

Output:

1/2340*(585*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(f*x + e))) 
) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(f*x + e)))) - sqrt 
(2)*log(sqrt(2)*sqrt(tan(f*x + e)) + tan(f*x + e) + 1) + sqrt(2)*log(-sqrt 
(2)*sqrt(tan(f*x + e)) + tan(f*x + e) + 1))/b^(5/2) + 8*(585*sqrt(b)/sqrt( 
tan(f*x + e)) - 117*sqrt(b)/tan(f*x + e)^(5/2) + 65*sqrt(b)/tan(f*x + e)^( 
9/2) - 45*sqrt(b)/tan(f*x + e)^(13/2))/b^3)/f
 

Giac [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx=\frac {1}{2340} \, b^{6} {\left (\frac {1170 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b^{10} f \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {1170 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b^{10} f \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} - \frac {585 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (f x + e\right ) + \sqrt {2} \sqrt {b \tan \left (f x + e\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b^{10} f \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {585 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (f x + e\right ) - \sqrt {2} \sqrt {b \tan \left (f x + e\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b^{10} f \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {8 \, {\left (585 \, b^{6} \tan \left (f x + e\right )^{6} - 117 \, b^{6} \tan \left (f x + e\right )^{4} + 65 \, b^{6} \tan \left (f x + e\right )^{2} - 45 \, b^{6}\right )}}{\sqrt {b \tan \left (f x + e\right )} b^{14} f \mathrm {sgn}\left (\tan \left (f x + e\right )\right ) \tan \left (f x + e\right )^{6}}\right )} \] Input:

integrate(1/(b*tan(f*x+e)^3)^(5/2),x, algorithm="giac")
 

Output:

1/2340*b^6*(1170*sqrt(2)*abs(b)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs 
(b)) + 2*sqrt(b*tan(f*x + e)))/sqrt(abs(b)))/(b^10*f*sgn(tan(f*x + e))) + 
1170*sqrt(2)*abs(b)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sq 
rt(b*tan(f*x + e)))/sqrt(abs(b)))/(b^10*f*sgn(tan(f*x + e))) - 585*sqrt(2) 
*abs(b)^(3/2)*log(b*tan(f*x + e) + sqrt(2)*sqrt(b*tan(f*x + e))*sqrt(abs(b 
)) + abs(b))/(b^10*f*sgn(tan(f*x + e))) + 585*sqrt(2)*abs(b)^(3/2)*log(b*t 
an(f*x + e) - sqrt(2)*sqrt(b*tan(f*x + e))*sqrt(abs(b)) + abs(b))/(b^10*f* 
sgn(tan(f*x + e))) + 8*(585*b^6*tan(f*x + e)^6 - 117*b^6*tan(f*x + e)^4 + 
65*b^6*tan(f*x + e)^2 - 45*b^6)/(sqrt(b*tan(f*x + e))*b^14*f*sgn(tan(f*x + 
 e))*tan(f*x + e)^6))
 

Mupad [F(-1)]

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

int(1/(b*tan(e + f*x)^3)^(5/2),x)
 

Output:

int(1/(b*tan(e + f*x)^3)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{8}}d x \right )}{b^{3}} \] Input:

int(1/(b*tan(f*x+e)^3)^(5/2),x)
 

Output:

(sqrt(b)*int(sqrt(tan(e + f*x))/tan(e + f*x)**8,x))/b**3