3.296 \(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=101 \[ -\frac{8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 a (5 A+3 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f} \]

[Out]

(-8*a^2*(5*A + 3*B)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(5*A + 3*B)*Cos[e + f*x]*Sqrt[a + a*S
in[e + f*x]])/(15*f) - (2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f)

________________________________________________________________________________________

Rubi [A]  time = 0.0868601, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2751, 2647, 2646} \[ -\frac{8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 a (5 A+3 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]),x]

[Out]

(-8*a^2*(5*A + 3*B)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(5*A + 3*B)*Cos[e + f*x]*Sqrt[a + a*S
in[e + f*x]])/(15*f) - (2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f)

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx &=-\frac{2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}+\frac{1}{5} (5 A+3 B) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac{2 a (5 A+3 B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}-\frac{2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}+\frac{1}{15} (4 a (5 A+3 B)) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a (5 A+3 B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}-\frac{2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}\\ \end{align*}

Mathematica [A]  time = 0.401855, size = 101, normalized size = 1. \[ -\frac{a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (2 (5 A+9 B) \sin (e+f x)+50 A-3 B \cos (2 (e+f x))+39 B)}{15 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]),x]

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(50*A + 39*B - 3*B*Cos[2*(e + f*x)] + 2*(
5*A + 9*B)*Sin[e + f*x]))/(15*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

________________________________________________________________________________________

Maple [A]  time = 1.089, size = 77, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ){a}^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \sin \left ( fx+e \right ) \left ( 5\,A+9\,B \right ) -3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+25\,A+21\,B \right ) }{15\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x)

[Out]

2/15*(1+sin(f*x+e))*a^2*(-1+sin(f*x+e))*(sin(f*x+e)*(5*A+9*B)-3*B*cos(f*x+e)^2+25*A+21*B)/cos(f*x+e)/(a+a*sin(
f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2), x)

________________________________________________________________________________________

Fricas [A]  time = 1.65723, size = 347, normalized size = 3.44 \begin{align*} \frac{2 \,{\left (3 \, B a \cos \left (f x + e\right )^{3} -{\left (5 \, A + 6 \, B\right )} a \cos \left (f x + e\right )^{2} -{\left (25 \, A + 21 \, B\right )} a \cos \left (f x + e\right ) - 4 \,{\left (5 \, A + 3 \, B\right )} a -{\left (3 \, B a \cos \left (f x + e\right )^{2} +{\left (5 \, A + 9 \, B\right )} a \cos \left (f x + e\right ) - 4 \,{\left (5 \, A + 3 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{15 \,{\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

2/15*(3*B*a*cos(f*x + e)^3 - (5*A + 6*B)*a*cos(f*x + e)^2 - (25*A + 21*B)*a*cos(f*x + e) - 4*(5*A + 3*B)*a - (
3*B*a*cos(f*x + e)^2 + (5*A + 9*B)*a*cos(f*x + e) - 4*(5*A + 3*B)*a)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f
*cos(f*x + e) + f*sin(f*x + e) + f)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}} \left (A + B \sin{\left (e + f x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*(A + B*sin(e + f*x)), x)

________________________________________________________________________________________

Giac [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((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out