3.167 \(\int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \, dx\)

Optimal. Leaf size=173 \[ \frac{904 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{128 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^4 \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x)}{11 d}+\frac{8 a^4 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{9 d}+\frac{150 a^4 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{77 d}+\frac{128 a^4 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{904 a^4 \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d} \]

[Out]

(128*a^4*EllipticE[(c + d*x)/2, 2])/(15*d) + (904*a^4*EllipticF[(c + d*x)/2, 2])/(231*d) + (904*a^4*Sqrt[Cos[c
 + d*x]]*Sin[c + d*x])/(231*d) + (128*a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (150*a^4*Cos[c + d*x]^(5/2
)*Sin[c + d*x])/(77*d) + (8*a^4*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d) + (2*a^4*Cos[c + d*x]^(9/2)*Sin[c + d*x
])/(11*d)

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Rubi [A]  time = 0.20459, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2757, 2635, 2641, 2639} \[ \frac{904 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{128 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^4 \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x)}{11 d}+\frac{8 a^4 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{9 d}+\frac{150 a^4 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{77 d}+\frac{128 a^4 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{904 a^4 \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^4,x]

[Out]

(128*a^4*EllipticE[(c + d*x)/2, 2])/(15*d) + (904*a^4*EllipticF[(c + d*x)/2, 2])/(231*d) + (904*a^4*Sqrt[Cos[c
 + d*x]]*Sin[c + d*x])/(231*d) + (128*a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (150*a^4*Cos[c + d*x]^(5/2
)*Sin[c + d*x])/(77*d) + (8*a^4*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d) + (2*a^4*Cos[c + d*x]^(9/2)*Sin[c + d*x
])/(11*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))^4 \, dx &=\int \left (a^4 \cos ^{\frac{3}{2}}(c+d x)+4 a^4 \cos ^{\frac{5}{2}}(c+d x)+6 a^4 \cos ^{\frac{7}{2}}(c+d x)+4 a^4 \cos ^{\frac{9}{2}}(c+d x)+a^4 \cos ^{\frac{11}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \cos ^{\frac{3}{2}}(c+d x) \, dx+a^4 \int \cos ^{\frac{11}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^{\frac{9}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^{\frac{7}{2}}(c+d x) \, dx\\ &=\frac{2 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{8 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{12 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{8 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{2 a^4 \cos ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{3} a^4 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{11} \left (9 a^4\right ) \int \cos ^{\frac{7}{2}}(c+d x) \, dx+\frac{1}{5} \left (12 a^4\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{9} \left (28 a^4\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx+\frac{1}{7} \left (30 a^4\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{24 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{74 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{128 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{150 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac{8 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{2 a^4 \cos ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{77} \left (45 a^4\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{7} \left (10 a^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{15} \left (28 a^4\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{128 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{74 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{904 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{128 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{150 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac{8 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{2 a^4 \cos ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{11 d}+\frac{1}{77} \left (15 a^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{128 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{904 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{904 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{128 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{150 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac{8 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{2 a^4 \cos ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [C]  time = 3.63354, size = 271, normalized size = 1.57 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (\frac{59136 \sec (c) \left (\csc (c) \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \left (3 \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+\cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )-2 \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )\right )}{\sqrt{\sec ^2(c)} \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}-108480 \sin (c) \sqrt{\csc ^2(c)} \cos (c+d x) \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\cos (c+d x) (122610 \sin (c+d x)+45584 \sin (2 (c+d x))+14445 \sin (3 (c+d x))+3080 \sin (4 (c+d x))+315 \sin (5 (c+d x))-236544 \cot (c))\right )}{443520 d \sqrt{\cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^4,x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(-108480*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[
c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + Cos
[c + d*x]*(-236544*Cot[c] + 122610*Sin[c + d*x] + 45584*Sin[2*(c + d*x)] + 14445*Sin[3*(c + d*x)] + 3080*Sin[4
*(c + d*x)] + 315*Sin[5*(c + d*x)]) + (59136*Sec[c]*(-2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTa
n[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Cs
c[c]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])))/(443520*d*Sqrt[C
os[c + d*x]])

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Maple [A]  time = 2.187, size = 273, normalized size = 1.6 \begin{align*} -{\frac{8\,{a}^{4}}{3465\,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 ( 5040\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{13}-5320\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}+1740\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+326\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+678\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-4465\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+1695\,\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 ) -3696\,\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 ) +2001\,\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)^4,x)

[Out]

-8/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(5040*cos(1/2*d*x+1/2*c)^13-5320*cos(1/2*d
*x+1/2*c)^11+1740*cos(1/2*d*x+1/2*c)^9+326*cos(1/2*d*x+1/2*c)^7+678*cos(1/2*d*x+1/2*c)^5-4465*cos(1/2*d*x+1/2*
c)^3+1695*(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))
-3696*(sin(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))+200
1*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 )}^{4} \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))^4,x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4*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^{4} \cos \left (d x + c\right )^{5} + 4 \, a^{4} \cos \left (d x + c\right )^{4} + 6 \, a^{4} \cos \left (d x + c\right )^{3} + 4 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4} \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))^4,x, algorithm="fricas")

[Out]

integral((a^4*cos(d*x + c)^5 + 4*a^4*cos(d*x + c)^4 + 6*a^4*cos(d*x + c)^3 + 4*a^4*cos(d*x + c)^2 + a^4*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))**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \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))^4,x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4*cos(d*x + c)^(3/2), x)