3.217 \(\int \frac{\sin ^4(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=100 \[ \frac{12 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d^3}-\frac{24 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{5 b d^2 \sqrt{\cos (a+b x)}}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}} \]

[Out]

(-24*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*d^2*Sqrt[Cos[a + b*x]]) + (12*(d*Cos[a + b*x])^(3/2)
*Sin[a + b*x])/(5*b*d^3) + (2*Sin[a + b*x]^3)/(b*d*Sqrt[d*Cos[a + b*x]])

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Rubi [A]  time = 0.104586, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2566, 2568, 2640, 2639} \[ \frac{12 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d^3}-\frac{24 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{5 b d^2 \sqrt{\cos (a+b x)}}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*d^2*Sqrt[Cos[a + b*x]]) + (12*(d*Cos[a + b*x])^(3/2)
*Sin[a + b*x])/(5*b*d^3) + (2*Sin[a + b*x]^3)/(b*d*Sqrt[d*Cos[a + b*x]])

Rule 2566

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

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 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 \frac{\sin ^4(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx &=\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{6 \int \sqrt{d \cos (a+b x)} \sin ^2(a+b x) \, dx}{d^2}\\ &=\frac{12 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d^3}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{12 \int \sqrt{d \cos (a+b x)} \, dx}{5 d^2}\\ &=\frac{12 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d^3}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}-\frac{\left (12 \sqrt{d \cos (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \, dx}{5 d^2 \sqrt{\cos (a+b x)}}\\ &=-\frac{24 \sqrt{d \cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b d^2 \sqrt{\cos (a+b x)}}+\frac{12 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d^3}+\frac{2 \sin ^3(a+b x)}{b d \sqrt{d \cos (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.0594919, size = 60, normalized size = 0.6 \[ \frac{\sin ^5(a+b x) \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac{5}{4},\frac{5}{2};\frac{7}{2};\sin ^2(a+b x)\right )}{5 b d \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.068, size = 213, normalized size = 2.1 \begin{align*} -{\frac{8}{5\,db}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}d+ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}d} \left ( -2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}\cos \left ( 1/2\,bx+a/2 \right ) +2\,\cos \left ( 1/2\,bx+a/2 \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+3\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) -3\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\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(sin(b*x+a)^4/(d*cos(b*x+a))^(3/2),x)

[Out]

-8/5/d*(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+2*
cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4+3*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*Ellipt
icE(cos(1/2*b*x+1/2*a),2^(1/2))-3*sin(1/2*b*x+1/2*a)^2*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 \frac{\sin \left (b x + a\right )^{4}}{\left (d \cos \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt{d \cos \left (b x + a\right )}}{d^{2} \cos \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral((cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sqrt(d*cos(b*x + a))/(d^2*cos(b*x + a)^2), 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(sin(b*x+a)**4/(d*cos(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{4}}{\left (d \cos \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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