3.1.81 \(\int \sin ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\) [81]

Optimal. Leaf size=98 \[ \frac {5 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{42 b}-\frac {5 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{42 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{14 b}-\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b} \]

[Out]

-5/42*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b-1/14*cos(2*b*x+2*a)
*sin(2*b*x+2*a)^(5/2)/b-1/18*sin(2*b*x+2*a)^(9/2)/b-5/42*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]
time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4383, 2715, 2720} \begin {gather*} -\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}+\frac {5 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{42 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{14 b}-\frac {5 \sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{42 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*EllipticF[a - Pi/4 + b*x, 2])/(42*b) - (5*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(42*b) - (Cos[2*a + 2*b*
x]*Sin[2*a + 2*b*x]^(5/2))/(14*b) - Sin[2*a + 2*b*x]^(9/2)/(18*b)

Rule 2715

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] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 4383

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-e^2)*(e*Sin
[a + b*x])^(m - 2)*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(m + 2*p))), x] + Dist[e^2*((m + p - 1)/(m + 2*p)), Int[(e
*Sin[a + b*x])^(m - 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && E
qQ[d/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && NeQ[m + 2*p, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \sin ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx &=-\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}+\frac {1}{2} \int \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\\ &=-\frac {\cos (2 a+2 b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{14 b}-\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}+\frac {5}{14} \int \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=-\frac {5 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{42 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{14 b}-\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}+\frac {5}{42} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {5 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{42 b}-\frac {5 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{42 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{14 b}-\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\\ \end {align*}

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Mathematica [A]
time = 3.15, size = 96, normalized size = 0.98 \begin {gather*} \frac {240 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {\sin (2 (a+b x))}-70 \sin (2 (a+b x))-156 \sin (4 (a+b x))+35 \sin (6 (a+b x))+18 \sin (8 (a+b x))-7 \sin (10 (a+b x))}{2016 b \sqrt {\sin (2 (a+b x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(240*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*(a + b*x)]] - 70*Sin[2*(a + b*x)] - 156*Sin[4*(a + b*x)] + 35*Sin
[6*(a + b*x)] + 18*Sin[8*(a + b*x)] - 7*Sin[10*(a + b*x)])/(2016*b*Sqrt[Sin[2*(a + b*x)]])

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 233.51, size = 519395265, normalized size = 5299951.68

method	result	size
default	\(\text {Expression too large to display}\)	\(519395265\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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