3.202 \(\int \frac{\sin ^2(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\)

Optimal. Leaf size=100 \[ -\frac{4 \sin (a+b x)}{5 b d^3 \sqrt{d \cos (a+b x)}}+\frac{4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{5 b d^4 \sqrt{\cos (a+b x)}}+\frac{2 \sin (a+b x)}{5 b d (d \cos (a+b x))^{5/2}} \]

[Out]

(4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*d^4*Sqrt[Cos[a + b*x]]) + (2*Sin[a + b*x])/(5*b*d*(d*C
os[a + b*x])^(5/2)) - (4*Sin[a + b*x])/(5*b*d^3*Sqrt[d*Cos[a + b*x]])

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Rubi [A]  time = 0.0817502, 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, 2636, 2640, 2639} \[ -\frac{4 \sin (a+b x)}{5 b d^3 \sqrt{d \cos (a+b x)}}+\frac{4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{d \cos (a+b x)}}{5 b d^4 \sqrt{\cos (a+b x)}}+\frac{2 \sin (a+b x)}{5 b d (d \cos (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(d*Cos[a + b*x])^(7/2),x]

[Out]

(4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*d^4*Sqrt[Cos[a + b*x]]) + (2*Sin[a + b*x])/(5*b*d*(d*C
os[a + b*x])^(5/2)) - (4*Sin[a + b*x])/(5*b*d^3*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 2636

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

Mathematica [C]  time = 0.0776264, size = 59, normalized size = 0.59 \[ \frac{\sin ^3(2 (a+b x)) \sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac{3}{2},\frac{9}{4};\frac{5}{2};\sin ^2(a+b x)\right )}{24 b (d \cos (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/(d*Cos[a + b*x])^(7/2),x]

[Out]

((Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[3/2, 9/4, 5/2, Sin[a + b*x]^2]*Sin[2*(a + b*x)]^3)/(24*b*(d*Cos[a +
b*x])^(7/2))

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Maple [B]  time = 0.091, size = 365, normalized size = 3.7 \begin{align*} -{\frac{4}{5\,{d}^{4}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 ( 4\,\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 ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}\cos \left ( 1/2\,bx+a/2 \right ) -4\,\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 ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+8\,\cos \left ( 1/2\,bx+a/2 \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) ,\sqrt{2} \right ) - \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}\cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) \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 ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-3} \left ( 8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}-12\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+6\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \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)^2/(d*cos(b*x+a))^(7/2),x)

[Out]

-4/5*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/d^4/sin(1/2*b*x+1/2*a)^3/(8*sin(1/2*b*x+1/2*a)^
6-12*sin(1/2*b*x+1/2*a)^4+6*sin(1/2*b*x+1/2*a)^2-1)*(4*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)
^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^4-8*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)-4*
(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*
x+1/2*a)^2+8*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4+(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(
1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))-sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a))*(-2*sin(1/2*b*x+1/2*a)^4*d
+sin(1/2*b*x+1/2*a)^2*d)^(1/2)/(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 )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

integral(-sqrt(d*cos(b*x + a))*(cos(b*x + a)^2 - 1)/(d^4*cos(b*x + a)^4), 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)**2/(d*cos(b*x+a))**(7/2),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 )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^2/(d*cos(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^2/(d*cos(b*x + a))^(7/2), x)