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

Optimal. Leaf size=121 \[ \frac{8 a^2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{4 a^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{8 a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{7 d} \]

[Out]

(12*a^2*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^2*EllipticF[(c + d*x)/2, 2])/(7*d) + (8*a^2*Sqrt[Cos[c + d*x]]
*Sin[c + d*x])/(7*d) + (4*a^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a^2*Cos[c + d*x]^(5/2)*Sin[c + d*x])
/(7*d)

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Rubi [A]  time = 0.118619, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2757, 2635, 2641, 2639} \[ \frac{8 a^2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{4 a^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{8 a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^2*EllipticF[(c + d*x)/2, 2])/(7*d) + (8*a^2*Sqrt[Cos[c + d*x]]
*Sin[c + d*x])/(7*d) + (4*a^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a^2*Cos[c + d*x]^(5/2)*Sin[c + d*x])
/(7*d)

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \, dx &=\int \left (a^2 \cos ^{\frac{3}{2}}(c+d x)+2 a^2 \cos ^{\frac{5}{2}}(c+d x)+a^2 \cos ^{\frac{7}{2}}(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^{\frac{3}{2}}(c+d x) \, dx+a^2 \int \cos ^{\frac{7}{2}}(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{4 a^2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{3} a^2 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{7} \left (5 a^2\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{5} \left (6 a^2\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{8 a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d}+\frac{4 a^2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{21} \left (5 a^2\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{12 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{8 a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d}+\frac{4 a^2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [C]  time = 6.11489, size = 500, normalized size = 4.13 \[ -\frac{3 \csc (c) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^2 \left (\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{2 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{10 d}-\frac{2 \csc (c) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^2 \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{7 d \sqrt{\cot ^2(c)+1}}+\sqrt{\cos (c+d x)} \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^2 \left (\frac{17 \sin (c) \cos (d x)}{56 d}+\frac{\sin (2 c) \cos (2 d x)}{10 d}+\frac{\sin (3 c) \cos (3 d x)}{56 d}+\frac{17 \cos (c) \sin (d x)}{56 d}+\frac{\cos (2 c) \sin (2 d x)}{10 d}+\frac{\cos (3 c) \sin (3 d x)}{56 d}-\frac{3 \cot (c)}{5 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*((-3*Cot[c])/(5*d) + (17*Cos[d*x]*Sin[c])/(56*d
) + (Cos[2*d*x]*Sin[2*c])/(10*d) + (Cos[3*d*x]*Sin[3*c])/(56*d) + (17*Cos[c]*Sin[d*x])/(56*d) + (Cos[2*c]*Sin[
2*d*x])/(10*d) + (Cos[3*c]*Sin[3*d*x])/(56*d)) - (2*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}
, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*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 + Sin[d*x - ArcTan[Cot[c]]]])/(
7*d*Sqrt[1 + Cot[c]^2]) - (3*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((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]]]]
*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 + Tan[c]^2]]))/(10*d)

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Maple [A]  time = 2.161, size = 272, normalized size = 2.3 \begin{align*} -{\frac{4\,{a}^{2}}{35\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 40\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-116\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +126\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +10\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -21\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -39\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/35*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-
116*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+126*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+10*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21*(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))-39*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+
1/2*c))/(-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{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \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(cos(d*x+c)^(3/2)*(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^2*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 ({\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*cos(d*x + c)^3 + 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*sqrt(cos(d*x + c)), 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(cos(d*x+c)**(3/2)*(a+a*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [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)^(3/2)*(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out