3.1166 \(\int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=285 \[ -\frac{(5 A+13 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

[Out]

((8*A + 19*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/
(4*a^(3/2)*d) - ((5*A + 13*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*
x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^
(7/2)*(a + a*Sec[c + d*x])^(3/2)) + ((A + 2*C)*Sin[c + d*x])/(2*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]
]) - ((2*A + 7*C)*Sin[c + d*x])/(4*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.910585, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.216, Rules used = {4265, 4085, 4021, 4023, 3808, 206, 3801, 215} \[ -\frac{(5 A+13 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

((8*A + 19*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/
(4*a^(3/2)*d) - ((5*A + 13*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*
x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^
(7/2)*(a + a*Sec[c + d*x])^(3/2)) + ((A + 2*C)*Sin[c + d*x])/(2*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]
]) - ((2*A + 7*C)*Sin[c + d*x])/(4*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4085

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

Rule 4021

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

Rule 4023

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

Rule 3808

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

Rule 206

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

Rule 3801

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

Rule 215

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

Rubi steps

\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (\frac{1}{2} a (A+5 C)-2 a (A+2 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-3 a^2 (A+2 C)+a^2 (2 A+7 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^3}\\ &=-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} a^3 (2 A+7 C)-\frac{1}{2} a^3 (8 A+19 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left ((5 A+13 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}+\frac{\left ((8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{8 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{\left ((5 A+13 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{2 a d}-\frac{\left ((8 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 a^2 d}\\ &=\frac{(8 A+19 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{4 a^{3/2} d}-\frac{(5 A+13 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.76652, size = 213, normalized size = 0.75 \[ -\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (A \cos ^2(c+d x)+C\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right ) ((2 A+7 C) \cos (2 (c+d x))+2 A+6 C \cos (c+d x)+3 C)+(5 A+13 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{3}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\frac{(8 A+19 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{3}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{2}}\right )}{4 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

-((C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]*((5*A + 13*C)*ArcTanh[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(
c + d*x))/2])^2 - ((8*A + 19*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2])^2)
/Sqrt[2] + (2*A + 3*C + 6*C*Cos[c + d*x] + (2*A + 7*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(4*a*d*Cos[c + d*x
]^(5/2)*(A + 2*C + A*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.3, size = 508, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/8/d*(-1+cos(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(8*A*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-8*A*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(co
s(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)+19*C*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos
(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-19*C*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(
d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)+4*A*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3-20*A*arctan(1
/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)+14*C*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3-5
2*C*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-4*A*cos(d*x+c)^2*(-2/(cos(d*x+c)+
1))^(1/2)-8*C*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)-10*C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+4*C*(-2/(cos(d*
x+c)+1))^(1/2))/a^2/sin(d*x+c)^3/(-2/(cos(d*x+c)+1))^(1/2)/cos(d*x+c)^(3/2)

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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((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.869465, size = 1956, normalized size = 6.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(2*sqrt(2)*((5*A + 13*C)*cos(d*x + c)^4 + 2*(5*A + 13*C)*cos(d*x + c)^3 + (5*A + 13*C)*cos(d*x + c)^2)*s
qrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*s
in(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((2*A + 7*C)*cos(d*x + c)^2 +
3*C*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 19*
C)*cos(d*x + c)^4 + 2*(8*A + 19*C)*cos(d*x + c)^3 + (8*A + 19*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3
- 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*
cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2
*d*cos(d*x + c)^2), 1/8*(2*sqrt(2)*((5*A + 13*C)*cos(d*x + c)^4 + 2*(5*A + 13*C)*cos(d*x + c)^3 + (5*A + 13*C)
*cos(d*x + c)^2)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(
a*sin(d*x + c))) - 2*((2*A + 7*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 19*C)*cos(d*x + c)^4 + 2*(8*A + 19*C)*cos(d*x + c)^3 + (8*A + 1
9*C)*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin
(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*c
os(d*x + c)^2)]

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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((A+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2)), x)