3.530 \(\int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=157 \[ -\frac{8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{105 f}-\frac{4 d (7 c-d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac{2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

[Out]

(-8*a^2*(35*c^2 + 42*c*d + 19*d^2)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(35*c^2 + 42*c*d + 19
*d^2)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) - (4*(7*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))
/(35*f) - (2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*a*f)

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Rubi [A]  time = 0.229463, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2761, 2751, 2647, 2646} \[ -\frac{8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{105 f}-\frac{4 d (7 c-d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac{2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*(35*c^2 + 42*c*d + 19*d^2)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(35*c^2 + 42*c*d + 19
*d^2)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) - (4*(7*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))
/(35*f) - (2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*a*f)

Rule 2761

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(
d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], 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, -1]

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} (c+d \sin (e+f x))^2 \, dx &=-\frac{2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac{2 \int (a+a \sin (e+f x))^{3/2} \left (\frac{1}{2} a \left (7 c^2+5 d^2\right )+a (7 c-d) d \sin (e+f x)\right ) \, dx}{7 a}\\ &=-\frac{4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac{2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac{1}{35} \left (35 c^2+42 c d+19 d^2\right ) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac{2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{105 f}-\frac{4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac{2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac{1}{105} \left (4 a \left (35 c^2+42 c d+19 d^2\right )\right ) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{8 a^2 \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x)}{105 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a \left (35 c^2+42 c d+19 d^2\right ) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{105 f}-\frac{4 (7 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac{2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}\\ \end{align*}

Mathematica [A]  time = 0.875155, size = 136, normalized size = 0.87 \[ -\frac{a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\left (140 c^2+504 c d+253 d^2\right ) \sin (e+f x)+700 c^2-6 d (14 c+13 d) \cos (2 (e+f x))+1092 c d-15 d^2 \sin (3 (e+f x))+494 d^2\right )}{210 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(700*c^2 + 1092*c*d + 494*d^2 - 6*d*(14*c
+ 13*d)*Cos[2*(e + f*x)] + (140*c^2 + 504*c*d + 253*d^2)*Sin[e + f*x] - 15*d^2*Sin[3*(e + f*x)]))/(210*f*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 0.668, size = 130, 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 ( 15\,{d}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}+42\,cd \left ( \sin \left ( fx+e \right ) \right ) ^{2}+39\,{d}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}+35\,{c}^{2}\sin \left ( fx+e \right ) +126\,\sin \left ( fx+e \right ) cd+52\,\sin \left ( fx+e \right ){d}^{2}+175\,{c}^{2}+252\,cd+104\,{d}^{2} \right ) }{105\,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)*(c+d*sin(f*x+e))^2,x)

[Out]

2/105*(1+sin(f*x+e))*a^2*(-1+sin(f*x+e))*(15*d^2*sin(f*x+e)^3+42*c*d*sin(f*x+e)^2+39*d^2*sin(f*x+e)^2+35*c^2*s
in(f*x+e)+126*sin(f*x+e)*c*d+52*sin(f*x+e)*d^2+175*c^2+252*c*d+104*d^2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^2, x)

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Fricas [A]  time = 1.62153, size = 590, normalized size = 3.76 \begin{align*} \frac{2 \,{\left (15 \, a d^{2} \cos \left (f x + e\right )^{4} + 3 \,{\left (14 \, a c d + 13 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 140 \, a c^{2} - 168 \, a c d - 76 \, a d^{2} -{\left (35 \, a c^{2} + 84 \, a c d + 43 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (175 \, a c^{2} + 294 \, a c d + 143 \, a d^{2}\right )} \cos \left (f x + e\right ) +{\left (15 \, a d^{2} \cos \left (f x + e\right )^{3} + 140 \, a c^{2} + 168 \, a c d + 76 \, a d^{2} - 6 \,{\left (7 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (35 \, a c^{2} + 126 \, a c d + 67 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{105 \,{\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)*(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

2/105*(15*a*d^2*cos(f*x + e)^4 + 3*(14*a*c*d + 13*a*d^2)*cos(f*x + e)^3 - 140*a*c^2 - 168*a*c*d - 76*a*d^2 - (
35*a*c^2 + 84*a*c*d + 43*a*d^2)*cos(f*x + e)^2 - (175*a*c^2 + 294*a*c*d + 143*a*d^2)*cos(f*x + e) + (15*a*d^2*
cos(f*x + e)^3 + 140*a*c^2 + 168*a*c*d + 76*a*d^2 - 6*(7*a*c*d + 4*a*d^2)*cos(f*x + e)^2 - (35*a*c^2 + 126*a*c
*d + 67*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*cos(f*x + e) + f*sin(f*x + e) + f)

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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 (c + d \sin{\left (e + f x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(c+d*sin(f*x+e))**2,x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*(c + d*sin(e + f*x))**2, x)

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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)*(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out