3.1247 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=266 \[ \frac{5 (19 A+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{48 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(299 A+27 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{48 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(163 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((163*A + 19*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[
c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((299*A + 27*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(48*a^
2*d*Sqrt[a + a*Cos[c + d*x]]) - ((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (
(17*A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + (5*(19*A + 3*C)*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(48*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.917349, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {4221, 3042, 2978, 2984, 12, 2782, 205} \[ \frac{5 (19 A+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{48 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(299 A+27 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{48 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(163 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((163*A + 19*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[
c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((299*A + 27*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(48*a^
2*d*Sqrt[a + a*Cos[c + d*x]]) - ((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (
(17*A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + (5*(19*A + 3*C)*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(48*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 4221

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

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2978

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

Rule 2984

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (11 A+3 C)-a (3 A-C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{4} a^2 (19 A+3 C)-a^2 (17 A+C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{5 (19 A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a^3 (299 A+27 C)+\frac{5}{4} a^3 (19 A+3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(299 A+27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{5 (19 A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{3 a^4 (163 A+19 C)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{6 a^6}\\ &=-\frac{(299 A+27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{5 (19 A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((163 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(299 A+27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{5 (19 A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((163 A+19 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac{(163 A+19 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{16 \sqrt{2} a^{5/2} d}-\frac{(299 A+27 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{5 (19 A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.43765, size = 243, normalized size = 0.91 \[ \frac{i \cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (3 (163 A+19 C) e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\frac{1}{8} i \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((1537 A+81 C) \cos (c+d x)+2 (503 A+39 C) \cos (2 (c+d x))+299 A \cos (3 (c+d x))+878 A+27 C \cos (3 (c+d x))+78 C)\right )}{12 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/12)*Cos[(c + d*x)/2]^5*((3*(163*A + 19*C)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I
)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/E^((I/2)*(c + d*x)) + (I
/8)*(878*A + 78*C + (1537*A + 81*C)*Cos[c + d*x] + 2*(503*A + 39*C)*Cos[2*(c + d*x)] + 299*A*Cos[3*(c + d*x)]
+ 27*C*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*Sec[c + d*x]^(3/2)*Tan[(c + d*x)/2]))/(d*(a*(1 + Cos[c + d*x]))^(5
/2))

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Maple [B]  time = 0.206, size = 573, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96/d*2^(1/2)/a^3*(-489*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/si
n(d*x+c))-57*C*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/sin(d*x+c))-14
67*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))/sin(d*x+c))-171*C*(cos(d
*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))/sin(d*x+c))-1467*A*(cos(d*x+c)/(1+c
os(d*x+c)))^(3/2)*sin(d*x+c)*cos(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-171*C*(cos(d*x+c)/(1+cos(d*x+c)))^(
3/2)*sin(d*x+c)*cos(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+299*A*2^(1/2)*cos(d*x+c)^4-489*A*(cos(d*x+c)/(1+
cos(d*x+c)))^(3/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+27*C*2^(1/2)*cos(d*x+c)^4-57*C*(cos(d*x+c)/(1
+cos(d*x+c)))^(3/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+204*A*cos(d*x+c)^3*2^(1/2)+12*C*cos(d*x+c)^3
*2^(1/2)-343*A*cos(d*x+c)^2*2^(1/2)-39*C*cos(d*x+c)^2*2^(1/2)-192*A*cos(d*x+c)*2^(1/2)+32*A*2^(1/2))*cos(d*x+c
)*(1/cos(d*x+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/(1+cos(d*x+c))^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*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.85552, size = 657, normalized size = 2.47 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left ({\left (299 \, A + 27 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (503 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 160 \, A \cos \left (d x + c\right ) - 32 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(3*sqrt(2)*((163*A + 19*C)*cos(d*x + c)^4 + 3*(163*A + 19*C)*cos(d*x + c)^3 + 3*(163*A + 19*C)*cos(d*x +
c)^2 + (163*A + 19*C)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(
a)*sin(d*x + c))) + 2*((299*A + 27*C)*cos(d*x + c)^3 + (503*A + 39*C)*cos(d*x + c)^2 + 160*A*cos(d*x + c) - 32
*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(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(-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*cos(d*x+c)**2)*sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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