\(\int \frac {1}{(a-a \sec ^2(c+d x))^{5/2}} \, dx\) [226]

Optimal result

Integrand size = 17, antiderivative size = 100 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {\cot (c+d x)}{2 a^2 d \sqrt {-a \tan ^2(c+d x)}}-\frac {\cot ^3(c+d x)}{4 a^2 d \sqrt {-a \tan ^2(c+d x)}}+\frac {\log (\sin (c+d x)) \tan (c+d x)}{a^2 d \sqrt {-a \tan ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {\cot (c+d x) \left (-4 \csc ^2(c+d x)+\csc ^4(c+d x)-4 \log (\sin (c+d x))\right ) \sqrt {-a \tan ^2(c+d x)}}{4 a^3 d} \] Input:

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

Output:

(Cot[c + d*x]*(-4*Csc[c + d*x]^2 + Csc[c + d*x]^4 - 4*Log[Sin[c + d*x]])*S 
qrt[-(a*Tan[c + d*x]^2)])/(4*a^3*d)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.71, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.882, Rules used = {3042, 4609, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4609

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4141

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

Int[(a - a*Sec[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])/(a^2*Sqrt[-(a*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]])
 

rule 4609

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A 
ctivateTrig[u*(b*tan[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ 
[a + b, 0]
 

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {\left (\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sin \left (d x +c \right )^{4}-\ln \left (\frac {2}{\cos \left (d x +c \right )+1}\right ) \sin \left (d x +c \right )^{4}-\frac {13 \cos \left (d x +c \right )^{4}}{32}-\frac {3 \cos \left (d x +c \right )^{2}}{16}+\frac {11}{32}\right ) \sec \left (d x +c \right ) \csc \left (d x +c \right )^{3}}{d \sqrt {-a \tan \left (d x +c \right )^{2}}\, a^{2}}\)	\(105\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) x}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}}-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (d x +c \right )}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {4 i \left ({\mathrm e}^{6 i \left (d x +c \right )}-{\mathrm e}^{4 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\)	\(298\)

Input:

int(1/(a-sec(d*x+c)^2*a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(ln(csc(d*x+c)-cot(d*x+c))*sin(d*x+c)^4-ln(2/(cos(d*x+c)+1))*sin(d*x+c 
)^4-13/32*cos(d*x+c)^4-3/16*cos(d*x+c)^2+11/32)/(-a*tan(d*x+c)^2)^(1/2)/a^ 
2*sec(d*x+c)*csc(d*x+c)^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {{\left (4 \, \cos \left (d x + c\right )^{3} - 4 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 3 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \] Input:

integrate(1/(a-a*sec(d*x+c)^2)^(5/2),x, algorithm="fricas")
 

Output:

1/4*(4*cos(d*x + c)^3 - 4*(cos(d*x + c)^5 - 2*cos(d*x + c)^3 + cos(d*x + c 
))*log(1/2*sin(d*x + c)) - 3*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a)/cos 
(d*x + c)^2)/((a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)*sin( 
d*x + c))
 

Sympy [F]

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

integrate(1/(a-a*sec(d*x+c)**2)**(5/2),x)
 

Output:

Integral((-a*sec(c + d*x)**2 + a)**(-5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt {-a} a^{2}} - \frac {4 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt {-a} a^{2}} + \frac {2 \, \sqrt {-a} \tan \left (d x + c\right )^{2} - \sqrt {-a}}{a^{3} \tan \left (d x + c\right )^{4}}}{4 \, d} \] Input:

integrate(1/(a-a*sec(d*x+c)^2)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {\frac {64 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{\sqrt {-a} a^{2}} - \frac {32 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )}{\sqrt {-a} a^{2}} - \frac {\sqrt {-a} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, \sqrt {-a} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{5}} + \frac {48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}{\sqrt {-a} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{64 \, d} \] Input:

integrate(1/(a-a*sec(d*x+c)^2)^(5/2),x, algorithm="giac")
 

Output:

1/64*(64*log(tan(1/2*d*x + 1/2*c)^2 + 1)/(sqrt(-a)*a^2) - 32*log(tan(1/2*d 
*x + 1/2*c)^2)/(sqrt(-a)*a^2) - (sqrt(-a)*a^2*tan(1/2*d*x + 1/2*c)^4 - 12* 
sqrt(-a)*a^2*tan(1/2*d*x + 1/2*c)^2)/a^5 + (48*tan(1/2*d*x + 1/2*c)^4 - 12 
*tan(1/2*d*x + 1/2*c)^2 + 1)/(sqrt(-a)*a^2*tan(1/2*d*x + 1/2*c)^4))/d
 

Mupad [F(-1)]

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

int(1/(a - a/cos(c + d*x)^2)^(5/2),x)
 

Output:

int(1/(a - a/cos(c + d*x)^2)^(5/2), x)
 

Reduce [F]

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

int(1/(a-a*sec(d*x+c)^2)^(5/2),x)
 

Output:

( - sqrt(a)*int(sqrt( - sec(c + d*x)**2 + 1)/(sec(c + d*x)**6 - 3*sec(c + 
d*x)**4 + 3*sec(c + d*x)**2 - 1),x))/a**3