3.193 \(\int \frac{(a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=210 \[ -\frac{a^3 (64 A+15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (8 A+5 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{5 d \sqrt{\cos (c+d x)}}+\frac{5 a^{5/2} C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}+\frac{2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(5*a^(5/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^3*(64*A + 15*C)*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(8*A + 5*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(5
*d*Sqrt[Cos[c + d*x]]) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*A*(a +
a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))

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Rubi [A]  time = 0.725002, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {3044, 2975, 2981, 2774, 216} \[ -\frac{a^3 (64 A+15 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (8 A+5 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{5 d \sqrt{\cos (c+d x)}}+\frac{5 a^{5/2} C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}+\frac{2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

(5*a^(5/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^3*(64*A + 15*C)*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(8*A + 5*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(5
*d*Sqrt[Cos[c + d*x]]) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*A*(a +
a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))

Rule 3044

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[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

Rule 2774

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

Rule 216

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

Rubi steps

\begin{align*} \int \frac{(a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^{5/2} \left (\frac{5 a A}{2}-\frac{1}{2} a (2 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{3}{4} a^2 (8 A+5 C)-\frac{1}{4} a^2 (16 A-15 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{2 a^2 (8 A+5 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{8} a^3 (32 A+45 C)-\frac{1}{8} a^3 (64 A+15 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=-\frac{a^3 (64 A+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (8 A+5 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{2} \left (5 a^2 C\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{a^3 (64 A+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (8 A+5 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{\left (5 a^2 C\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{5 a^{5/2} C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}-\frac{a^3 (64 A+15 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (8 A+5 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.89289, size = 141, normalized size = 0.67 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) ((112 A+45 C) \cos (c+d x)+4 (43 A+15 C) \cos (2 (c+d x))+196 A+15 C \cos (3 (c+d x))+60 C)+300 \sqrt{2} C \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^{\frac{5}{2}}(c+d x)\right )}{120 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(300*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^
(5/2) + 2*(196*A + 60*C + (112*A + 45*C)*Cos[c + d*x] + 4*(43*A + 15*C)*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x
)])*Sin[(c + d*x)/2]))/(120*d*Cos[c + d*x]^(5/2))

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Maple [A]  time = 0.161, size = 245, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2}}{15\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( -75\,C\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-75\,C\sin \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \cos \left ( dx+c \right ) +15\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+86\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+15\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}-58\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-30\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-22\,A\cos \left ( dx+c \right ) -6\,A \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)

[Out]

-1/15/d*a^2*(a*(1+cos(d*x+c)))^(1/2)*(-75*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)^2-75*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arct
an(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)+15*C*cos(d*x+c)^4+86*A*cos(d*x+c)^3+15*
C*cos(d*x+c)^3-58*A*cos(d*x+c)^2-30*C*cos(d*x+c)^2-22*A*cos(d*x+c)-6*A)/sin(d*x+c)/cos(d*x+c)^(5/2)

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Maxima [B]  time = 2.20763, size = 1520, normalized size = 7.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/60*(15*(2*(a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (a^2*cos(d*x + c) - a
^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2
*cos(2*d*x + 2*c) + 1)*sqrt(a) + 5*(a^2*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - a^2*arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c
os(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x +
c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos
(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c) + 1)) + 1) + a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2
*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 8*(a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)
)*sin(d*x + c) - (a^2*cos(d*x + c) - a^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a))*C
/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) + 32*(15*sqrt(2)*a^(5/2)*sin(d*x + c
)/(cos(d*x + c) + 1) - 35*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 28*sqrt(2)*a^(5/2)*sin(d*x + c
)^5/(cos(d*x + c) + 1)^5 - 8*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*A/((sin(d*x + c)/(cos(d*x +
c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)))/d

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Fricas [A]  time = 1.70235, size = 451, normalized size = 2.15 \begin{align*} \frac{{\left (15 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (43 \, A + 15 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, A a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 75 \,{\left (C a^{2} \cos \left (d x + c\right )^{4} + C a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \arctan \left (\frac{\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 )}{15 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/15*((15*C*a^2*cos(d*x + c)^3 + 2*(43*A + 15*C)*a^2*cos(d*x + c)^2 + 28*A*a^2*cos(d*x + c) + 6*A*a^2)*sqrt(a*
cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 75*(C*a^2*cos(d*x + c)^4 + C*a^2*cos(d*x + c)^3)*sqrt(a)*a
rctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^
3)

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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+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),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 )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(7/2), x)