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

Optimal. Leaf size=194 \[ \frac{2 a \left (7 a^2+15 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 b \left (27 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 b \left (27 a^2+7 b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{2 a \left (7 a^2+15 b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 b^2 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}{9 d}+\frac{40 a b^2 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d} \]

[Out]

(2*b*(27*a^2 + 7*b^2)*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*a*(7*a^2 + 15*b^2)*EllipticF[(c + d*x)/2, 2])/(21
*d) + (2*a*(7*a^2 + 15*b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*b*(27*a^2 + 7*b^2)*Cos[c + d*x]^(3/2)
*Sin[c + d*x])/(45*d) + (40*a*b^2*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*b^2*Cos[c + d*x]^(5/2)*(a + b*C
os[c + d*x])*Sin[c + d*x])/(9*d)

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Rubi [A]  time = 0.22039, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2793, 3023, 2748, 2635, 2641, 2639} \[ \frac{2 a \left (7 a^2+15 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 b \left (27 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 b \left (27 a^2+7 b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{2 a \left (7 a^2+15 b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 b^2 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}{9 d}+\frac{40 a b^2 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(27*a^2 + 7*b^2)*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*a*(7*a^2 + 15*b^2)*EllipticF[(c + d*x)/2, 2])/(21
*d) + (2*a*(7*a^2 + 15*b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*b*(27*a^2 + 7*b^2)*Cos[c + d*x]^(3/2)
*Sin[c + d*x])/(45*d) + (40*a*b^2*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*b^2*Cos[c + d*x]^(5/2)*(a + b*C
os[c + d*x])*Sin[c + d*x])/(9*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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 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+b \cos (c+d x))^3 \, dx &=\frac{2 b^2 \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{9 d}+\frac{2}{9} \int \cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} a \left (9 a^2+5 b^2\right )+\frac{1}{2} b \left (27 a^2+7 b^2\right ) \cos (c+d x)+10 a b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{40 a b^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 b^2 \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{9 d}+\frac{4}{63} \int \cos ^{\frac{3}{2}}(c+d x) \left (\frac{9}{4} a \left (7 a^2+15 b^2\right )+\frac{7}{4} b \left (27 a^2+7 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{40 a b^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 b^2 \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{9} \left (b \left (27 a^2+7 b^2\right )\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx+\frac{1}{7} \left (a \left (7 a^2+15 b^2\right )\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{2 a \left (7 a^2+15 b^2\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 b \left (27 a^2+7 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{40 a b^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 b^2 \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{9 d}+\frac{1}{15} \left (b \left (27 a^2+7 b^2\right )\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (a \left (7 a^2+15 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 b \left (27 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a \left (7 a^2+15 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \left (7 a^2+15 b^2\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 b \left (27 a^2+7 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{40 a b^2 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac{2 b^2 \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.91535, size = 137, normalized size = 0.71 \[ \frac{60 \left (7 a^3+15 a b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+84 \left (27 a^2 b+7 b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} \left (7 b \left (108 a^2+43 b^2\right ) \cos (c+d x)+5 \left (84 a^3+54 a b^2 \cos (2 (c+d x))+234 a b^2+7 b^3 \cos (3 (c+d x))\right )\right )}{630 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(84*(27*a^2*b + 7*b^3)*EllipticE[(c + d*x)/2, 2] + 60*(7*a^3 + 15*a*b^2)*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[
c + d*x]]*(7*b*(108*a^2 + 43*b^2)*Cos[c + d*x] + 5*(84*a^3 + 234*a*b^2 + 54*a*b^2*Cos[2*(c + d*x)] + 7*b^3*Cos
[3*(c + d*x)]))*Sin[c + d*x])/(630*d)

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Maple [B]  time = 3.132, size = 470, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^10+(2160*a*b^2+2240*b^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1512*a^2*b-3240*a*b^2-2072*b^3)*sin(1/2*d
*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(420*a^3+1512*a^2*b+2520*a*b^2+952*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+(-210*a^3-378*a^2*b-720*a*b^2-168*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+105*(sin(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))*a^3+225*(sin(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))*a*b^2-567*(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))*a^2*b-147*(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))*b^3)/(-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 (b \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(cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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(cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

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