3.134 \(\int \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ \frac{163 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{95 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}-\frac{197 \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{17 \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

(163*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Sec[c + d*x]
^3*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - (17*Sec[c + d*x]^2*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*
x])^(3/2)) - (197*Tan[c + d*x])/(24*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + (95*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x
])/(48*a^3*d)

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Rubi [A]  time = 0.423982, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3816, 4019, 4010, 4001, 3795, 203} \[ \frac{163 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{95 \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}-\frac{197 \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{17 \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(163*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Sec[c + d*x]
^3*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - (17*Sec[c + d*x]^2*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*
x])^(3/2)) - (197*Tan[c + d*x])/(24*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + (95*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x
])/(48*a^3*d)

Rule 3816

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(2*m + 1)), x] + Dist[d^2/(a*b*(2*m + 1)), In
t[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m]
)

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int \frac{\sec ^3(c+d x) \left (3 a-\frac{11}{2} a \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{17 \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec ^2(c+d x) \left (17 a^2-\frac{95}{4} a^2 \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{17 \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{95 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}-\frac{\int \frac{\sec (c+d x) \left (-\frac{95 a^3}{8}+\frac{197}{4} a^3 \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{17 \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{197 \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{95 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}+\frac{163 \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{17 \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{197 \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{95 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}-\frac{163 \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{163 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{17 \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{197 \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{95 \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}\\ \end{align*}

Mathematica [A]  time = 1.25889, size = 135, normalized size = 0.74 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left (32 \sec ^3(c+d x)-160 \sec ^2(c+d x)-503 \sec (c+d x)-299\right )+978 \sqrt{2} \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{48 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((978*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 + Sqrt[1 - Sec[c + d*x
]]*(-299 - 503*Sec[c + d*x] - 160*Sec[c + d*x]^2 + 32*Sec[c + d*x]^3))*Tan[c + d*x])/(48*d*Sqrt[1 - Sec[c + d*
x]]*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.179, size = 417, normalized size = 2.3 \begin{align*} -{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{192\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) } \left ( 489\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1467\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1467\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +489\,\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -1196\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-816\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1372\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+768\,\cos \left ( dx+c \right ) -128 \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192/d/a^3*(-1+cos(d*x+c))^2*(489*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+
c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^3*sin(d*x+c)+1467*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)*cos(d*x+c)^2+1467*ln(((-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*c
os(d*x+c)*sin(d*x+c)+489*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)-1196*cos(d*x+c)^4-816*cos(d*x+c)^3+1372*cos(d*x+c)^2+768*cos(d*x+c)-1
28)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)

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Maxima [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(sec(d*x+c)^5/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.05429, size = 1220, normalized size = 6.67 \begin{align*} \left [-\frac{489 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \log \left (\frac{2 \, \sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}, -\frac{489 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) + 2 \,{\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{96 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/192*(489*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*log((2*sqr
t(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos
(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(299*cos(d*x + c)^3 + 503*cos(d*x + c)^2 + 160*cos(d
*x + c) - 32)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x +
c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/96*(489*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3
*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(s
qrt(a)*sin(d*x + c))) + 2*(299*cos(d*x + c)^3 + 503*cos(d*x + c)^2 + 160*cos(d*x + c) - 32)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 +
a^3*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 10.284, size = 343, normalized size = 1.87 \begin{align*} -\frac{\frac{{\left ({\left (3 \,{\left (\frac{2 \, \sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{23 \, \sqrt{2}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{668 \, \sqrt{2}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{465 \, \sqrt{2}}{a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{489 \, \sqrt{2} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(((3*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/(a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 23*sqrt(2)/(a*sgn(tan(1/2*d
*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 - 668*sqrt(2)/(a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1
/2*c)^2 + 465*sqrt(2)/(a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a
)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 489*sqrt(2)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d