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

Optimal. Leaf size=87 \[ \frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0666751, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2748, 2635, 2641, 2639} \[ \frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 5.35802, size = 232, normalized size = 2.67 \[ \frac{a (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-18 \cos (c) \sqrt{\sec ^2(c)} \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \csc \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )-20 \sin (c) \sqrt{\csc ^2(c)} \cos (c+d x) \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+2 \cos (c+d x) (10 \sin (c+d x)+3 \sin (2 (c+d x))-18 \cot (c))+\frac{9 \csc (c) \sec (c) \left (3 \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+\cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\sec ^2(c)}}\right )}{60 d \sqrt{\cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*((9*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])
*Csc[c]*Sec[c])/Sqrt[Sec[c]^2] - 20*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*Hypergeometr
icPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 2*Cos[c + d*x]*(-18*C
ot[c] + 10*Sin[c + d*x] + 3*Sin[2*(c + d*x)]) - 18*Cos[c]*Csc[d*x + ArcTan[Tan[c]]]*HypergeometricPFQ[{-1/2, -
1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(60*d*Sqrt[Cos[c
+ d*x]])

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Maple [A]  time = 2.177, size = 219, normalized size = 2.5 \begin{align*} -{\frac{2\,a}{15\,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 ( 24\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-28\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+5\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -9\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +4\,\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),x)

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(24*cos(1/2*d*x+1/2*c)^7-28*cos(1/2*d*x+1/2*c)
^5+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+4*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 )} \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)),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)*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 \cos \left (d x + c\right )^{2} + a \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)),x, algorithm="fricas")

[Out]

integral((a*cos(d*x + c)^2 + a*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)),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)),x, algorithm="giac")

[Out]

Timed out