3.140 \(\int \frac{(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=169 \[ \frac{4 a^3 (5 A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 (5 A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{4 a^3 (5 A-6 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}-\frac{2 (5 A-B) \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}+\frac{2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt{\cos (c+d x)}} \]

[Out]

(4*a^3*(5*A + 9*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(5*A + 3*B)*EllipticF[(c + d*x)/2, 2])/(3*d) - (4
*a^3*(5*A - 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt
[Cos[c + d*x]]) - (2*(5*A - B)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.434165, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2975, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac{4 a^3 (5 A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 (5 A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{4 a^3 (5 A-6 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}-\frac{2 (5 A-B) \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{5 d}+\frac{2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(5*A + 9*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(5*A + 3*B)*EllipticF[(c + d*x)/2, 2])/(3*d) - (4
*a^3*(5*A - 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt
[Cos[c + d*x]]) - (2*(5*A - B)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(5*d)

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 2976

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

Rule 2968

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+2 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{2} a (5 A+B)-\frac{1}{2} a (5 A-B) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{2 (5 A-B) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{4}{5} \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{2} a^2 (10 A+3 B)-\frac{1}{2} a^2 (5 A-6 B) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{2 (5 A-B) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{4}{5} \int \frac{\frac{1}{2} a^3 (10 A+3 B)+\left (-\frac{1}{2} a^3 (5 A-6 B)+\frac{1}{2} a^3 (10 A+3 B)\right ) \cos (c+d x)-\frac{1}{2} a^3 (5 A-6 B) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a^3 (5 A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{2 (5 A-B) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{8}{15} \int \frac{\frac{5}{4} a^3 (5 A+3 B)+\frac{3}{4} a^3 (5 A+9 B) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a^3 (5 A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{2 (5 A-B) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{1}{3} \left (2 a^3 (5 A+3 B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (2 a^3 (5 A+9 B)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 (5 A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (5 A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^3 (5 A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-\frac{2 (5 A-B) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 6.41195, size = 888, normalized size = 5.25 \[ \sqrt{\cos (c+d x)} (\cos (c+d x) a+a)^3 \left (-\frac{(15 \cos (2 c) A+5 A+18 B+18 B \cos (2 c)) \csc (c) \sec (c)}{40 d}+\frac{A \sec (c+d x) \sin (d x) \sec (c)}{4 d}+\frac{(A+3 B) \cos (d x) \sin (c)}{12 d}+\frac{B \cos (2 d x) \sin (2 c)}{40 d}+\frac{(A+3 B) \cos (c) \sin (d x)}{12 d}+\frac{B \cos (2 c) \sin (2 d x)}{40 d}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )-\frac{A (\cos (c+d x) a+a)^3 \csc (c) \left (\frac{\, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt{\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}} \sqrt{\tan ^2(c)+1}}-\frac{\frac{2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac{\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{\tan ^2(c)+1}}}{\sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}}}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{4 d}-\frac{9 B (\cos (c+d x) a+a)^3 \csc (c) \left (\frac{\, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt{\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}} \sqrt{\tan ^2(c)+1}}-\frac{\frac{2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac{\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{\tan ^2(c)+1}}}{\sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}}}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{20 d}-\frac{5 A (\cos (c+d x) a+a)^3 \csc (c) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{6 d \sqrt{\cot ^2(c)+1}}-\frac{B (\cos (c+d x) a+a)^3 \csc (c) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d \sqrt{\cot ^2(c)+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-((5*A + 18*B + 15*A*Cos[2*c] + 18*B*Cos[2*c])
*Csc[c]*Sec[c])/(40*d) + ((A + 3*B)*Cos[d*x]*Sin[c])/(12*d) + (B*Cos[2*d*x]*Sin[2*c])/(40*d) + ((A + 3*B)*Cos[
c]*Sin[d*x])/(12*d) + (A*Sec[c]*Sec[c + d*x]*Sin[d*x])/(4*d) + (B*Cos[2*c]*Sin[2*d*x])/(40*d)) - (5*A*(a + a*C
os[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*S
ec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan
[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(6*d*Sqrt[1 + Cot[c]^2]) - (B*(a + a*Cos[c + d*x])^3*Csc[c]*H
ypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]
]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Si
n[d*x - ArcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) - (A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((H
ypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 -
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 +
Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x
+ ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + T
an[c]^2]]))/(4*d) - (9*B*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4},
{3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqr
t[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2])
- ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan
[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(20*d)

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Maple [B]  time = 4.093, size = 519, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*a^3*(-12*B*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A+21*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*A+9*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*A*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-15*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(si
n(1/2*d*x+1/2*c)^2)^(1/2)-27*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*
x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B a^{3} \cos \left (d x + c\right )^{4} +{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + A a^{3}}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*a^3*cos(d*x + c)^4 + (A + 3*B)*a^3*cos(d*x + c)^3 + 3*(A + B)*a^3*cos(d*x + c)^2 + (3*A + B)*a^3*c
os(d*x + c) + A*a^3)/cos(d*x + c)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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