\(\int \frac {1}{(b \tan ^2(c+d x))^{5/2}} \, dx\) [29]

Optimal result

Integrand size = 14, antiderivative size = 97 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\cot (c+d x)}{2 b^2 d \sqrt {b \tan ^2(c+d x)}}-\frac {\cot ^3(c+d x)}{4 b^2 d \sqrt {b \tan ^2(c+d x)}}+\frac {\log (\sin (c+d x)) \tan (c+d x)}{b^2 d \sqrt {b \tan ^2(c+d x)}} \] Output:

1/2*cot(d*x+c)/b^2/d/(b*tan(d*x+c)^2)^(1/2)-1/4*cot(d*x+c)^3/b^2/d/(b*tan( 
d*x+c)^2)^(1/2)+ln(sin(d*x+c))*tan(d*x+c)/b^2/d/(b*tan(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\cot (c+d x) \left (-4 \csc ^2(c+d x)+\csc ^4(c+d x)-4 \log (\sin (c+d x))\right ) \sqrt {b \tan ^2(c+d x)}}{4 b^3 d} \] Input:

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

Output:

-1/4*(Cot[c + d*x]*(-4*Csc[c + d*x]^2 + Csc[c + d*x]^4 - 4*Log[Sin[c + d*x 
]])*Sqrt[b*Tan[c + d*x]^2])/(b^3*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 4141, 3042, 25, 3954, 25, 3042, 25, 3954, 25, 3042, 25, 3956}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4141

\(\displaystyle \frac {\tan (c+d x) \int \cot ^5(c+d x)dx}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\tan (c+d x) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^5dx}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\tan (c+d x) \left (\int -\tan \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\cot ^4(c+d x)}{4 d}\right )}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3954

\(\displaystyle -\frac {\tan (c+d x) \left (\int -\cot (c+d x)dx+\frac {\cot ^4(c+d x)}{4 d}-\frac {\cot ^2(c+d x)}{2 d}\right )}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\tan (c+d x) \left (-\int \cot (c+d x)dx+\frac {\cot ^4(c+d x)}{4 d}-\frac {\cot ^2(c+d x)}{2 d}\right )}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\tan (c+d x) \left (-\int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\cot ^4(c+d x)}{4 d}-\frac {\cot ^2(c+d x)}{2 d}\right )}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\tan (c+d x) \left (\int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx+\frac {\cot ^4(c+d x)}{4 d}-\frac {\cot ^2(c+d x)}{2 d}\right )}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

\(\Big \downarrow \) 3956

\(\displaystyle -\frac {\tan (c+d x) \left (\frac {\cot ^4(c+d x)}{4 d}-\frac {\cot ^2(c+d x)}{2 d}-\frac {\log (-\sin (c+d x))}{d}\right )}{b^2 \sqrt {b \tan ^2(c+d x)}}\)

Input:

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

Output:

-(((-1/2*Cot[c + d*x]^2/d + Cot[c + d*x]^4/(4*d) - Log[-Sin[c + d*x]]/d)*T 
an[c + d*x])/(b^2*Sqrt[b*Tan[c + d*x]^2]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3954

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

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

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 = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right ) \left (4 \ln \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{4}-2 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{4}+2 \tan \left (d x +c \right )^{2}-1\right )}{4 d \left (b \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}\)	\(74\)
default	\(\frac {\tan \left (d x +c \right ) \left (4 \ln \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{4}-2 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{4}+2 \tan \left (d x +c \right )^{2}-1\right )}{4 d \left (b \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}\)	\(74\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) x}{b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}}-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (d x +c \right )}{b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {4 i \left ({\mathrm e}^{6 i \left (d x +c \right )}-{\mathrm e}^{4 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\)	\(302\)

Input:

int(1/(b*tan(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*tan(d*x+c)*(4*ln(tan(d*x+c))*tan(d*x+c)^4-2*ln(1+tan(d*x+c)^2)*tan(d 
*x+c)^4+2*tan(d*x+c)^2-1)/(b*tan(d*x+c)^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {{\left (2 \, \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} - 1\right )} \sqrt {b \tan \left (d x + c\right )^{2}}}{4 \, b^{3} d \tan \left (d x + c\right )^{5}} \] Input:

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

Output:

1/4*(2*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 3*tan(d*x 
 + c)^4 + 2*tan(d*x + c)^2 - 1)*sqrt(b*tan(d*x + c)^2)/(b^3*d*tan(d*x + c) 
^5)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

Integral((b*tan(c + d*x)**2)**(-5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{b^{\frac {5}{2}}} - \frac {4 \, \log \left (\tan \left (d x + c\right )\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sqrt {b} \tan \left (d x + c\right )^{2} - \sqrt {b}}{b^{3} \tan \left (d x + c\right )^{4}}}{4 \, d} \] Input:

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

Output:

-1/4*(2*log(tan(d*x + c)^2 + 1)/b^(5/2) - 4*log(tan(d*x + c))/b^(5/2) - (2 
*sqrt(b)*tan(d*x + c)^2 - sqrt(b))/(b^3*tan(d*x + c)^4))/d
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{b d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} - \frac {2 \, \log \left (\tan \left (d x + c\right )^{2}\right )}{b d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} + \frac {3 \, \tan \left (d x + c\right )^{4} - 2 \, \tan \left (d x + c\right )^{2} + 1}{b d \mathrm {sgn}\left (\tan \left (d x + c\right )\right ) \tan \left (d x + c\right )^{4}}}{4 \, b^{\frac {3}{2}}} \] Input:

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

Output:

-1/4*(2*log(tan(d*x + c)^2 + 1)/(b*d*sgn(tan(d*x + c))) - 2*log(tan(d*x + 
c)^2)/(b*d*sgn(tan(d*x + c))) + (3*tan(d*x + c)^4 - 2*tan(d*x + c)^2 + 1)/ 
(b*d*sgn(tan(d*x + c))*tan(d*x + c)^4))/b^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{5/2}} \, dx=\frac {\sqrt {b}\, \left (-2 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right )^{4}+4 \,\mathrm {log}\left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{4}+2 \tan \left (d x +c \right )^{2}-1\right )}{4 \tan \left (d x +c \right )^{4} b^{3} d} \] Input:

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

Output:

(sqrt(b)*( - 2*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**4 + 4*log(tan(c + d* 
x))*tan(c + d*x)**4 + 2*tan(c + d*x)**2 - 1))/(4*tan(c + d*x)**4*b**3*d)