3.1161 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=268 \[ -\frac{(15 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A+5 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(13 A+5 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{10 a d \sqrt{a \sec (c+d x)+a}}+\frac{(49 A+25 C) \sin (c+d x)}{10 a d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((15*A + 7*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)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Se
c[c + d*x])^(3/2)) + ((49*A + 25*C)*Sin[c + d*x])/(10*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((13*
A + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A + 5*C)*Cos[c + d*x]^(3/2)*
Sin[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.865277, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4265, 4085, 4022, 4013, 3808, 206} \[ -\frac{(15 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A+5 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(13 A+5 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{10 a d \sqrt{a \sec (c+d x)+a}}+\frac{(49 A+25 C) \sin (c+d x)}{10 a d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*A + 7*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)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Se
c[c + d*x])^(3/2)) + ((49*A + 25*C)*Sin[c + d*x])/(10*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((13*
A + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A + 5*C)*Cos[c + d*x]^(3/2)*
Sin[c + d*x])/(10*a*d*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 4022

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

Rule 4013

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

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])

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (9 A+5 C)+a (3 A+C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(9 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{4} a^2 (13 A+5 C)-a^2 (9 A+5 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(13 A+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{8} a^3 (49 A+25 C)+\frac{3}{4} a^3 (13 A+5 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(49 A+25 C) \sin (c+d x)}{10 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(13 A+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}-\frac{\left ((15 A+7 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{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(49 A+25 C) \sin (c+d x)}{10 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(13 A+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{\left ((15 A+7 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{(15 A+7 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) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(49 A+25 C) \sin (c+d x)}{10 a d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(13 A+5 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}+\frac{(9 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.1879, size = 118, normalized size = 0.44 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) ((39 A+20 C) \cos (c+d x)-2 A \cos (2 (c+d x))+A \cos (3 (c+d x))+47 A+25 C)-5 (15 A+7 C) \cos \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{10 a d \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(15*A + 7*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + (47*A + 25*C + (39*A + 20*C)*Cos[c + d*x] - 2*A*
Cos[2*(c + d*x)] + A*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(10*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])]
)

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Maple [A]  time = 0.375, size = 318, normalized size = 1.2 \begin{align*}{\frac{-1+\cos \left ( dx+c \right ) }{20\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-75\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-35\,C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-16\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-75\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}A\sin \left ( dx+c \right ) -35\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sin \left ( dx+c \right ) +80\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+40\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+26\,A\cos \left ( dx+c \right ) +10\,C\cos \left ( dx+c \right ) -98\,A-50\,C \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(8*A*cos(d*x+c)^4-75*A*sin(d*x+c)*
cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-35*C*cos(d*x+c)*sin(d*x+
c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-16*A*cos(d*x+c)^3-75*arctan(1/2*
sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*A*sin(d*x+c)-35*C*(-2/(cos(d*x+c)+1))^(1/2)*ar
ctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+80*A*cos(d*x+c)^2+40*C*cos(d*x+c)^2+26*A*cos(d*x+c)+
10*C*cos(d*x+c)-98*A-50*C)/a^2/sin(d*x+c)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.557679, size = 1234, normalized size = 4.6 \begin{align*} \left [\frac{5 \, \sqrt{2}{\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 7 \, C\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left (4 \, A \cos \left (d x + c\right )^{3} - 4 \, A \cos \left (d x + c\right )^{2} + 4 \,{\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 49 \, A + 25 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{40 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac{5 \, \sqrt{2}{\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 7 \, C\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \,{\left (4 \, A \cos \left (d x + c\right )^{3} - 4 \, A \cos \left (d x + c\right )^{2} + 4 \,{\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 49 \, A + 25 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{20 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^2 + 2*(15*A + 7*C)*cos(d*x + c) + 15*A + 7*C)*sqrt(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))*sin(d*x + c) - 2*a*c
os(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(4*A*cos(d*x + c)^3 - 4*A*cos(d*x + c)^2 + 4*(9*
A + 5*C)*cos(d*x + c) + 49*A + 25*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/
(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/20*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^2 + 2*(15*A
+ 7*C)*cos(d*x + c) + 15*A + 7*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqr
t(cos(d*x + c))/(a*sin(d*x + c))) + 2*(4*A*cos(d*x + c)^3 - 4*A*cos(d*x + c)^2 + 4*(9*A + 5*C)*cos(d*x + c) +
49*A + 25*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 +
2*a^2*d*cos(d*x + c) + a^2*d)]

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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(cos(d*x+c)**(5/2)*(A+C*sec(d*x+c)**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{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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