\(\int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) [83]

Optimal result

Integrand size = 22, antiderivative size = 69 \[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}-\frac {\cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{6 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b} \]

[Out]

-1/6*(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-1/10*sin(2*b*x+2*a)^
(5/2)/b-1/6*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4383, 2715, 2720} \[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{6 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{6 b} \]

[In]

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

[Out]

EllipticF[a - Pi/4 + b*x, 2]/(6*b) - (Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(6*b) - Sin[2*a + 2*b*x]^(5/2)/
(10*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*} \text {integral}& = -\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {1}{2} \int \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx \\ & = -\frac {\cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{6 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {1}{6} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}-\frac {\cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{6 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {20 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 (a+b x))}-9 \sin (2 (a+b x))-10 \sin (4 (a+b x))+3 \sin (6 (a+b x))}{120 b \sqrt {\sin (2 (a+b x))}} \]

[In]

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

[Out]

(20*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*(a + b*x)]] - 9*Sin[2*(a + b*x)] - 10*Sin[4*(a + b*x)] + 3*Sin[6*(
a + b*x)])/(120*b*Sqrt[Sin[2*(a + b*x)]])

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 65.32 (sec) , antiderivative size = 185748620, normalized size of antiderivative = 2692008.99

method	result	size
default	\(\text {Expression too large to display}\)	\(185748620\)

[In]

int(sin(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(sin(b*x+a)**2*sin(2*b*x+2*a)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sin ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2} \,d x \]

[In]

int(sin(a + b*x)^2*sin(2*a + 2*b*x)^(3/2),x)

[Out]

int(sin(a + b*x)^2*sin(2*a + 2*b*x)^(3/2), x)