3.213 \(\int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx\)

Optimal. Leaf size=128 \[ \frac{8 d^2 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{65 b \sqrt{\cos (a+b x)}}-\frac{2 \sin ^3(a+b x) (d \cos (a+b x))^{7/2}}{13 b d}-\frac{4 \sin (a+b x) (d \cos (a+b x))^{7/2}}{39 b d}+\frac{8 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{195 b} \]

[Out]

(8*d^2*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(65*b*Sqrt[Cos[a + b*x]]) + (8*d*(d*Cos[a + b*x])^(3/2)
*Sin[a + b*x])/(195*b) - (4*(d*Cos[a + b*x])^(7/2)*Sin[a + b*x])/(39*b*d) - (2*(d*Cos[a + b*x])^(7/2)*Sin[a +
b*x]^3)/(13*b*d)

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Rubi [A]  time = 0.12575, antiderivative size = 128, 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 = {2568, 2635, 2640, 2639} \[ \frac{8 d^2 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{65 b \sqrt{\cos (a+b x)}}-\frac{2 \sin ^3(a+b x) (d \cos (a+b x))^{7/2}}{13 b d}-\frac{4 \sin (a+b x) (d \cos (a+b x))^{7/2}}{39 b d}+\frac{8 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{195 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(5/2)*Sin[a + b*x]^4,x]

[Out]

(8*d^2*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(65*b*Sqrt[Cos[a + b*x]]) + (8*d*(d*Cos[a + b*x])^(3/2)
*Sin[a + b*x])/(195*b) - (4*(d*Cos[a + b*x])^(7/2)*Sin[a + b*x])/(39*b*d) - (2*(d*Cos[a + b*x])^(7/2)*Sin[a +
b*x]^3)/(13*b*d)

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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 (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx &=-\frac{2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac{6}{13} \int (d \cos (a+b x))^{5/2} \sin ^2(a+b x) \, dx\\ &=-\frac{4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac{2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac{4}{39} \int (d \cos (a+b x))^{5/2} \, dx\\ &=\frac{8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac{4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac{2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac{1}{65} \left (4 d^2\right ) \int \sqrt{d \cos (a+b x)} \, dx\\ &=\frac{8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac{4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac{2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac{\left (4 d^2 \sqrt{d \cos (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{65 \sqrt{\cos (a+b x)}}\\ &=\frac{8 d^2 \sqrt{d \cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{65 b \sqrt{\cos (a+b x)}}+\frac{8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac{4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac{2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}\\ \end{align*}

Mathematica [C]  time = 0.0735713, size = 65, normalized size = 0.51 \[ \frac{\sin ^2(a+b x) \sqrt [4]{\cos ^2(a+b x)} \tan ^3(a+b x) (d \cos (a+b x))^{5/2} \, _2F_1\left (-\frac{3}{4},\frac{5}{2};\frac{7}{2};\sin ^2(a+b x)\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(5/2)*Sin[a + b*x]^4,x]

[Out]

((d*Cos[a + b*x])^(5/2)*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-3/4, 5/2, 7/2, Sin[a + b*x]^2]*Sin[a + b*x]^
2*Tan[a + b*x]^3)/(5*b)

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Maple [A]  time = 0.06, size = 249, normalized size = 2. \begin{align*} -{\frac{8\,{d}^{3}}{195\,b}\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( 480\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{15}-1920\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{13}+3040\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{11}-2400\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{9}+958\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{7}-156\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{5}-5\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}-3\,\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) +3\,\cos \left ( 1/2\,bx+a/2 \right ) \right ){\frac{1}{\sqrt{-d \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2} \right ) }}} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{d \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x)

[Out]

-8/195*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^3*(480*cos(1/2*b*x+1/2*a)^15-1920*cos(1/2*b
*x+1/2*a)^13+3040*cos(1/2*b*x+1/2*a)^11-2400*cos(1/2*b*x+1/2*a)^9+958*cos(1/2*b*x+1/2*a)^7-156*cos(1/2*b*x+1/2
*a)^5-5*cos(1/2*b*x+1/2*a)^3-3*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(-2*cos(1/2*b*x+1/2*a)^2+1)^(1/2)*EllipticE(cos(1/
2*b*x+1/2*a),2^(1/2))+3*cos(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b
*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}} \sin \left (b x + a\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(5/2)*sin(b*x + a)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d^{2} \cos \left (b x + a\right )^{6} - 2 \, d^{2} \cos \left (b x + a\right )^{4} + d^{2} \cos \left (b x + a\right )^{2}\right )} \sqrt{d \cos \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

integral((d^2*cos(b*x + a)^6 - 2*d^2*cos(b*x + a)^4 + d^2*cos(b*x + a)^2)*sqrt(d*cos(b*x + a)), 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((d*cos(b*x+a))**(5/2)*sin(b*x+a)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}} \sin \left (b x + a\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(5/2)*sin(b*x + a)^4, x)