\(\int \sin ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\) [90]

Optimal result

Integrand size = 22, antiderivative size = 110 \[ \int \sin ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}+\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}-\frac {5 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b} \]

[Out]

-5/32*arcsin(cos(b*x+a)-sin(b*x+a))/b+5/32*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b-1/8*sin(b*x+a)*sin
(2*b*x+2*a)^(3/2)/b-5/16*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 110, 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, 4387, 4390} \[ \int \sin ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b}-\frac {5 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{16 b}+\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]

[In]

Int[Sin[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

(-5*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(32*b) + (5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(32*b) - (5*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(16*b) - (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(8*b)

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]

Rule 4387

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[-2*Cos[a + b*x]*((g*Sin[c
+ d*x])^p/(d*(2*p + 1))), x] + Dist[2*p*(g/(2*p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fr
eeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4390

Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-ArcSin[Cos[a + b*x] - Sin[a + b*
x]]/d, x] + Simp[Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[
b*c - a*d, 0] && EqQ[d/b, 2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b}+\frac {5}{8} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {5 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b}+\frac {5}{16} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}+\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}-\frac {5 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{8 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.78 \[ \int \sin ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {5 \left (-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+2 (-6 \cos (a+b x)+\cos (3 (a+b x))) \sqrt {\sin (2 (a+b x))}}{32 b} \]

[In]

Integrate[Sin[a + b*x]^3*Sqrt[Sin[2*a + 2*b*x]],x]

[Out]

(5*(-ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]) + 2*(-6*
Cos[a + b*x] + Cos[3*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/(32*b)

Maple [B] (warning: unable to verify)

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

Time = 16.22 (sec) , antiderivative size = 57690707, normalized size of antiderivative = 524460.97

method	result	size
default	\(\text {Expression too large to display}\)	\(57690707\)

[In]

int(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (96) = 192\).

Time = 0.26 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.55 \[ \int \sin ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - 9 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 10 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 10 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 5 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{128 \, b} \]

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="fricas")

[Out]

1/128*(8*sqrt(2)*(4*cos(b*x + a)^3 - 9*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 10*arctan(-(sqrt(2)*sqr
t(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(cos(b*x + a)^2 + 2*co
s(b*x + a)*sin(b*x + a) - 1)) - 10*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x
+ a))/(cos(b*x + a) - sin(b*x + a))) - 5*log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + a)^3 - (4*cos(b*x +
a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)
*sin(b*x + a) + 1))/b

Sympy [F(-1)]

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

[In]

integrate(sin(b*x+a)**3*sin(2*b*x+2*a)**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \sin ^3(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0]ext_reduce Error: Bad Argument TypeDone

Mupad [F(-1)]

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

[In]

int(sin(a + b*x)^3*sin(2*a + 2*b*x)^(1/2),x)

[Out]

int(sin(a + b*x)^3*sin(2*a + 2*b*x)^(1/2), x)