3.211 \(\int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx\)

Optimal. Leaf size=156 \[ \frac{56 d^3 \sin (a+b x) (d \cos (a+b x))^{3/2}}{3315 b}+\frac{56 d^4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{1105 b \sqrt{\cos (a+b x)}}-\frac{2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}-\frac{12 \sin (a+b x) (d \cos (a+b x))^{11/2}}{221 b d}+\frac{8 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{663 b} \]

[Out]

(56*d^4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(1105*b*Sqrt[Cos[a + b*x]]) + (56*d^3*(d*Cos[a + b*x])
^(3/2)*Sin[a + b*x])/(3315*b) + (8*d*(d*Cos[a + b*x])^(7/2)*Sin[a + b*x])/(663*b) - (12*(d*Cos[a + b*x])^(11/2
)*Sin[a + b*x])/(221*b*d) - (2*(d*Cos[a + b*x])^(11/2)*Sin[a + b*x]^3)/(17*b*d)

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Rubi [A]  time = 0.149011, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2568, 2635, 2640, 2639} \[ \frac{56 d^3 \sin (a+b x) (d \cos (a+b x))^{3/2}}{3315 b}+\frac{56 d^4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{1105 b \sqrt{\cos (a+b x)}}-\frac{2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}-\frac{12 \sin (a+b x) (d \cos (a+b x))^{11/2}}{221 b d}+\frac{8 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{663 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^(9/2)*Sin[a + b*x]^4,x]

[Out]

(56*d^4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(1105*b*Sqrt[Cos[a + b*x]]) + (56*d^3*(d*Cos[a + b*x])
^(3/2)*Sin[a + b*x])/(3315*b) + (8*d*(d*Cos[a + b*x])^(7/2)*Sin[a + b*x])/(663*b) - (12*(d*Cos[a + b*x])^(11/2
)*Sin[a + b*x])/(221*b*d) - (2*(d*Cos[a + b*x])^(11/2)*Sin[a + b*x]^3)/(17*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))^{9/2} \sin ^4(a+b x) \, dx &=-\frac{2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d}+\frac{6}{17} \int (d \cos (a+b x))^{9/2} \sin ^2(a+b x) \, dx\\ &=-\frac{12 (d \cos (a+b x))^{11/2} \sin (a+b x)}{221 b d}-\frac{2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d}+\frac{12}{221} \int (d \cos (a+b x))^{9/2} \, dx\\ &=\frac{8 d (d \cos (a+b x))^{7/2} \sin (a+b x)}{663 b}-\frac{12 (d \cos (a+b x))^{11/2} \sin (a+b x)}{221 b d}-\frac{2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d}+\frac{1}{663} \left (28 d^2\right ) \int (d \cos (a+b x))^{5/2} \, dx\\ &=\frac{56 d^3 (d \cos (a+b x))^{3/2} \sin (a+b x)}{3315 b}+\frac{8 d (d \cos (a+b x))^{7/2} \sin (a+b x)}{663 b}-\frac{12 (d \cos (a+b x))^{11/2} \sin (a+b x)}{221 b d}-\frac{2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d}+\frac{\left (28 d^4\right ) \int \sqrt{d \cos (a+b x)} \, dx}{1105}\\ &=\frac{56 d^3 (d \cos (a+b x))^{3/2} \sin (a+b x)}{3315 b}+\frac{8 d (d \cos (a+b x))^{7/2} \sin (a+b x)}{663 b}-\frac{12 (d \cos (a+b x))^{11/2} \sin (a+b x)}{221 b d}-\frac{2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d}+\frac{\left (28 d^4 \sqrt{d \cos (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{1105 \sqrt{\cos (a+b x)}}\\ &=\frac{56 d^4 \sqrt{d \cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{1105 b \sqrt{\cos (a+b x)}}+\frac{56 d^3 (d \cos (a+b x))^{3/2} \sin (a+b x)}{3315 b}+\frac{8 d (d \cos (a+b x))^{7/2} \sin (a+b x)}{663 b}-\frac{12 (d \cos (a+b x))^{11/2} \sin (a+b x)}{221 b d}-\frac{2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d}\\ \end{align*}

Mathematica [C]  time = 0.130511, size = 57, normalized size = 0.37 \[ \frac{\sqrt [4]{\cos ^2(a+b x)} \tan ^5(a+b x) (d \cos (a+b x))^{9/2} \, _2F_1\left (-\frac{7}{4},\frac{5}{2};\frac{7}{2};\sin ^2(a+b x)\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^(9/2)*Sin[a + b*x]^4,x]

[Out]

((d*Cos[a + b*x])^(9/2)*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-7/4, 5/2, 7/2, Sin[a + b*x]^2]*Tan[a + b*x]^
5)/(5*b)

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Maple [A]  time = 0.069, size = 275, normalized size = 1.8 \begin{align*} -{\frac{8\,{d}^{5}}{3315\,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 ( 24960\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{19}-124800\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{17}+265440\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{15}-312960\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{13}+222520\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{11}-96360\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{9}+23866\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{7}-2652\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{5}-35\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}-21\,\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 ) +21\,\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))^(9/2)*sin(b*x+a)^4,x)

[Out]

-8/3315*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^5*(24960*cos(1/2*b*x+1/2*a)^19-124800*cos(
1/2*b*x+1/2*a)^17+265440*cos(1/2*b*x+1/2*a)^15-312960*cos(1/2*b*x+1/2*a)^13+222520*cos(1/2*b*x+1/2*a)^11-96360
*cos(1/2*b*x+1/2*a)^9+23866*cos(1/2*b*x+1/2*a)^7-2652*cos(1/2*b*x+1/2*a)^5-35*cos(1/2*b*x+1/2*a)^3-21*(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))+21*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{9}{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))^(9/2)*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(9/2)*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^{4} \cos \left (b x + a\right )^{8} - 2 \, d^{4} \cos \left (b x + a\right )^{6} + d^{4} \cos \left (b x + a\right )^{4}\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))^(9/2)*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

integral((d^4*cos(b*x + a)^8 - 2*d^4*cos(b*x + a)^6 + d^4*cos(b*x + a)^4)*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))**(9/2)*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{9}{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))^(9/2)*sin(b*x+a)^4,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(9/2)*sin(b*x + a)^4, x)