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

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

[Out]

2/5*sin(b*x+a)/b/d/(d*cos(b*x+a))^(5/2)-4/5*sin(b*x+a)/b/d^3/(d*cos(b*x+a))^(1/2)+4/5*(cos(1/2*a+1/2*b*x)^2)^(
1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)/b/d^4/cos(b*x+a)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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]

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*}

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Mathematica [C]  time = 0.06, 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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} - 1\right )}}{d^{4} \cos \left (b x + a\right )^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.13, size = 365, normalized size = 3.65 \[ -\frac {4 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (4 \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \EllipticE \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-8 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-4 \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \EllipticE \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \EllipticE \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d}}{5 d^{4} \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \left (8 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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