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

Optimal result

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

[Out]

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*tan(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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4206, 3739, 3554, 3556} \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\cot ^3(c+d x)}{4 a^2 d \sqrt {-a \tan ^2(c+d x)}}+\frac {\cot (c+d x)}{2 a^2 d \sqrt {-a \tan ^2(c+d x)}}+\frac {\tan (c+d x) \log (\sin (c+d x))}{a^2 d \sqrt {-a \tan ^2(c+d x)}} \]

[In]

Int[(a - a*Sec[c + d*x]^2)^(-5/2),x]

[Out]

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

Rule 3554

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

Rule 3556

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

Rule 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[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 4206

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (-a \tan ^2(c+d x)\right )^{5/2}} \, dx \\ & = \frac {\tan (c+d x) \int \cot ^5(c+d x) \, dx}{a^2 \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 {\tan (c+d x) \int \cot ^3(c+d x) \, dx}{a^2 \sqrt {-a \tan ^2(c+d x)}} \\ & = \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 {\tan (c+d x) \int \cot (c+d x) \, dx}{a^2 \sqrt {-a \tan ^2(c+d x)}} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {2 \cot (c+d x)-\cot ^3(c+d x)+4 (\log (\cos (c+d x))+\log (\tan (c+d x))) \tan (c+d x)}{4 a^2 d \sqrt {-a \tan ^2(c+d x)}} \]

[In]

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

[Out]

(2*Cot[c + d*x] - Cot[c + d*x]^3 + 4*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])*Tan[c + d*x])/(4*a^2*d*Sqrt[-(a*T
an[c + d*x]^2)])

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\left (-32 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sin \left (d x +c \right )^{4}+32 \ln \left (\frac {2}{\cos \left (d x +c \right )+1}\right ) \sin \left (d x +c \right )^{4}+13 \cos \left (d x +c \right )^{4}+6 \cos \left (d x +c \right )^{2}-11\right ) \sec \left (d x +c \right ) \csc \left (d x +c \right )^{3}}{32 d \sqrt {-a \tan \left (d x +c \right )^{2}}\, a^{2}}\)	\(107\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) x}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}}-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (d x +c \right )}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {4 i \left ({\mathrm e}^{6 i \left (d x +c \right )}-{\mathrm e}^{4 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\)	\(298\)

[In]

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

[Out]

-1/32/d*(-32*ln(-cot(d*x+c)+csc(d*x+c))*sin(d*x+c)^4+32*ln(2/(cos(d*x+c)+1))*sin(d*x+c)^4+13*cos(d*x+c)^4+6*co
s(d*x+c)^2-11)/(-a*tan(d*x+c)^2)^(1/2)/a^2*sec(d*x+c)*csc(d*x+c)^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {{\left (4 \, \cos \left (d x + c\right )^{3} - 4 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 3 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

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)*si
n(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt {-a} a^{2}} - \frac {4 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt {-a} a^{2}} + \frac {2 \, \sqrt {-a} \tan \left (d x + c\right )^{2} - \sqrt {-a}}{a^{3} \tan \left (d x + c\right )^{4}}}{4 \, d} \]

[In]

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

[Out]

-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)

none

Time = 0.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{5/2}} \, dx=\frac {\frac {64 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{\sqrt {-a} a^{2}} - \frac {32 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )}{\sqrt {-a} a^{2}} - \frac {\sqrt {-a} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, \sqrt {-a} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{5}} + \frac {48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}{\sqrt {-a} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{64 \, d} \]

[In]

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

[Out]

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

[In]

int(1/(a - a/cos(c + d*x)^2)^(5/2),x)

[Out]

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