3.227 \(\int \frac{1}{(a-a \sec ^2(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=133 \[ \frac{\cot ^5(c+d x)}{6 a^3 d \sqrt{-a \tan ^2(c+d x)}}-\frac{\cot ^3(c+d x)}{4 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\cot (c+d x)}{2 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\tan (c+d x) \log (\sin (c+d x))}{a^3 d \sqrt{-a \tan ^2(c+d x)}} \]

[Out]

Cot[c + d*x]/(2*a^3*d*Sqrt[-(a*Tan[c + d*x]^2)]) - Cot[c + d*x]^3/(4*a^3*d*Sqrt[-(a*Tan[c + d*x]^2)]) + Cot[c
+ d*x]^5/(6*a^3*d*Sqrt[-(a*Tan[c + d*x]^2)]) + (Log[Sin[c + d*x]]*Tan[c + d*x])/(a^3*d*Sqrt[-(a*Tan[c + d*x]^2
)])

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Rubi [A]  time = 0.0612302, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {4121, 3658, 3473, 3475} \[ \frac{\cot ^5(c+d x)}{6 a^3 d \sqrt{-a \tan ^2(c+d x)}}-\frac{\cot ^3(c+d x)}{4 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\cot (c+d x)}{2 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\tan (c+d x) \log (\sin (c+d x))}{a^3 d \sqrt{-a \tan ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(-7/2),x]

[Out]

Cot[c + d*x]/(2*a^3*d*Sqrt[-(a*Tan[c + d*x]^2)]) - Cot[c + d*x]^3/(4*a^3*d*Sqrt[-(a*Tan[c + d*x]^2)]) + Cot[c
+ d*x]^5/(6*a^3*d*Sqrt[-(a*Tan[c + d*x]^2)]) + (Log[Sin[c + d*x]]*Tan[c + d*x])/(a^3*d*Sqrt[-(a*Tan[c + d*x]^2
)])

Rule 4121

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]

Rule 3658

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 3473

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 3475

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a-a \sec ^2(c+d x)\right )^{7/2}} \, dx &=\int \frac{1}{\left (-a \tan ^2(c+d x)\right )^{7/2}} \, dx\\ &=-\frac{\tan (c+d x) \int \cot ^7(c+d x) \, dx}{a^3 \sqrt{-a \tan ^2(c+d x)}}\\ &=\frac{\cot ^5(c+d x)}{6 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\tan (c+d x) \int \cot ^5(c+d x) \, dx}{a^3 \sqrt{-a \tan ^2(c+d x)}}\\ &=-\frac{\cot ^3(c+d x)}{4 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\cot ^5(c+d x)}{6 a^3 d \sqrt{-a \tan ^2(c+d x)}}-\frac{\tan (c+d x) \int \cot ^3(c+d x) \, dx}{a^3 \sqrt{-a \tan ^2(c+d x)}}\\ &=\frac{\cot (c+d x)}{2 a^3 d \sqrt{-a \tan ^2(c+d x)}}-\frac{\cot ^3(c+d x)}{4 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\cot ^5(c+d x)}{6 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\tan (c+d x) \int \cot (c+d x) \, dx}{a^3 \sqrt{-a \tan ^2(c+d x)}}\\ &=\frac{\cot (c+d x)}{2 a^3 d \sqrt{-a \tan ^2(c+d x)}}-\frac{\cot ^3(c+d x)}{4 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\cot ^5(c+d x)}{6 a^3 d \sqrt{-a \tan ^2(c+d x)}}+\frac{\log (\sin (c+d x)) \tan (c+d x)}{a^3 d \sqrt{-a \tan ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.337888, size = 79, normalized size = 0.59 \[ -\frac{\tan ^7(c+d x) \left (2 \cot ^6(c+d x)-3 \cot ^4(c+d x)+6 \cot ^2(c+d x)+12 \log (\tan (c+d x))+12 \log (\cos (c+d x))\right )}{12 d \left (-a \tan ^2(c+d x)\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sec[c + d*x]^2)^(-7/2),x]

[Out]

-((6*Cot[c + d*x]^2 - 3*Cot[c + d*x]^4 + 2*Cot[c + d*x]^6 + 12*Log[Cos[c + d*x]] + 12*Log[Tan[c + d*x]])*Tan[c
+ d*x]^7)/(12*d*(-(a*Tan[c + d*x]^2))^(7/2))

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Maple [B]  time = 0.261, size = 265, normalized size = 2. \begin{align*} -{\frac{\sin \left ( dx+c \right ) }{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{7}} \left ( 48\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\ln \left ( 2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) -48\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +25\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}-144\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ( 2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) +144\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+144\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\ln \left ( 2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) -144\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -33\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-48\,\ln \left ( 2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) +48\,\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +19 \right ) \left ( -{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*sec(d*x+c)^2)^(7/2),x)

[Out]

-1/48/d*(48*cos(d*x+c)^6*ln(2/(cos(d*x+c)+1))-48*cos(d*x+c)^6*ln(-(-1+cos(d*x+c))/sin(d*x+c))+25*cos(d*x+c)^6-
144*cos(d*x+c)^4*ln(2/(cos(d*x+c)+1))+144*cos(d*x+c)^4*ln(-(-1+cos(d*x+c))/sin(d*x+c))-3*cos(d*x+c)^4+144*cos(
d*x+c)^2*ln(2/(cos(d*x+c)+1))-144*cos(d*x+c)^2*ln(-(-1+cos(d*x+c))/sin(d*x+c))-33*cos(d*x+c)^2-48*ln(2/(cos(d*
x+c)+1))+48*ln(-(-1+cos(d*x+c))/sin(d*x+c))+19)*sin(d*x+c)/cos(d*x+c)^7/(-a*sin(d*x+c)^2/cos(d*x+c)^2)^(7/2)

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Maxima [A]  time = 1.49992, size = 127, normalized size = 0.95 \begin{align*} -\frac{\frac{6 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt{-a} a^{3}} - \frac{12 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt{-a} a^{3}} + \frac{6 \, \sqrt{-a} \tan \left (d x + c\right )^{4} - 3 \, \sqrt{-a} \tan \left (d x + c\right )^{2} + 2 \, \sqrt{-a}}{a^{4} \tan \left (d x + c\right )^{6}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^(7/2),x, algorithm="maxima")

[Out]

-1/12*(6*log(tan(d*x + c)^2 + 1)/(sqrt(-a)*a^3) - 12*log(tan(d*x + c))/(sqrt(-a)*a^3) + (6*sqrt(-a)*tan(d*x +
c)^4 - 3*sqrt(-a)*tan(d*x + c)^2 + 2*sqrt(-a))/(a^4*tan(d*x + c)^6))/d

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Fricas [A]  time = 0.525282, size = 406, normalized size = 3.05 \begin{align*} \frac{{\left (18 \, \cos \left (d x + c\right )^{5} - 27 \, \cos \left (d x + c\right )^{3} - 12 \,{\left (\cos \left (d x + c\right )^{7} - 3 \, \cos \left (d x + c\right )^{5} + 3 \, \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 ) + 11 \, \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}}}}{12 \,{\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^(7/2),x, algorithm="fricas")

[Out]

1/12*(18*cos(d*x + c)^5 - 27*cos(d*x + c)^3 - 12*(cos(d*x + c)^7 - 3*cos(d*x + c)^5 + 3*cos(d*x + c)^3 - cos(d
*x + c))*log(1/2*sin(d*x + c)) + 11*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a)/cos(d*x + c)^2)/((a^4*d*cos(d*x
+ c)^6 - 3*a^4*d*cos(d*x + c)^4 + 3*a^4*d*cos(d*x + c)^2 - a^4*d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)**2)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.96491, size = 383, normalized size = 2.88 \begin{align*} -\frac{\frac{384 \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}{\sqrt{-a} a^{3} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{192 \, \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )}{\sqrt{-a} a^{3} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} + \frac{352 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 87 \, \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^{3} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6}} + \frac{\sqrt{-a} a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 12 \, \sqrt{-a} a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 87 \, \sqrt{-a} a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{12} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^(7/2),x, algorithm="giac")

[Out]

-1/384*(384*log(tan(1/2*d*x + 1/2*c)^2 + 1)/(sqrt(-a)*a^3*sgn(-tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c)))
- 192*log(tan(1/2*d*x + 1/2*c)^2)/(sqrt(-a)*a^3*sgn(-tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))) + (352*t
an(1/2*d*x + 1/2*c)^6 - 87*tan(1/2*d*x + 1/2*c)^4 + 12*tan(1/2*d*x + 1/2*c)^2 - 1)/(sqrt(-a)*a^3*sgn(-tan(1/2*
d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^6) + (sqrt(-a)*a^8*tan(1/2*d*x + 1/2*c)^6 - 12*sqr
t(-a)*a^8*tan(1/2*d*x + 1/2*c)^4 + 87*sqrt(-a)*a^8*tan(1/2*d*x + 1/2*c)^2)/(a^12*sgn(-tan(1/2*d*x + 1/2*c)^3 +
tan(1/2*d*x + 1/2*c))))/d