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

Optimal. Leaf size=228 \[ \frac{a^{3/2} (c-11 d) (c+d)^2 \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{8 d^{3/2} f}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a \sin (e+f x)+a}}+\frac{a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt{a \sin (e+f x)+a}}+\frac{a^2 (c-11 d) (c+d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{8 d f \sqrt{a \sin (e+f x)+a}} \]

[Out]

(a^(3/2)*(c - 11*d)*(c + d)^2*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]])])/(8*d^(3/2)*f) + (a^2*(c - 11*d)*(c + d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(8*d*f*Sqrt[a + a*S
in[e + f*x]]) + (a^2*(c - 11*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(12*d*f*Sqrt[a + a*Sin[e + f*x]]) - (
a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(3*d*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.43496, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2763, 21, 2770, 2775, 205} \[ \frac{a^{3/2} (c-11 d) (c+d)^2 \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{8 d^{3/2} f}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a \sin (e+f x)+a}}+\frac{a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt{a \sin (e+f x)+a}}+\frac{a^2 (c-11 d) (c+d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{8 d f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/2)*(c - 11*d)*(c + d)^2*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]])])/(8*d^(3/2)*f) + (a^2*(c - 11*d)*(c + d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(8*d*f*Sqrt[a + a*S
in[e + f*x]]) + (a^2*(c - 11*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(12*d*f*Sqrt[a + a*Sin[e + f*x]]) - (
a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(3*d*f*Sqrt[a + a*Sin[e + f*x]])

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2770

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

Rule 2775

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2} \, dx &=-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a+a \sin (e+f x)}}+\frac{\int \frac{\left (-\frac{1}{2} a^2 (c-11 d)-\frac{1}{2} a^2 (c-11 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)}} \, dx}{3 d}\\ &=-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a+a \sin (e+f x)}}-\frac{(a (c-11 d)) \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2} \, dx}{6 d}\\ &=\frac{a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a+a \sin (e+f x)}}-\frac{(a (c-11 d) (c+d)) \int \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)} \, dx}{8 d}\\ &=\frac{a^2 (c-11 d) (c+d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{8 d f \sqrt{a+a \sin (e+f x)}}+\frac{a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a+a \sin (e+f x)}}-\frac{\left (a (c-11 d) (c+d)^2\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{16 d}\\ &=\frac{a^2 (c-11 d) (c+d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{8 d f \sqrt{a+a \sin (e+f x)}}+\frac{a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (c-11 d) (c+d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{8 d f}\\ &=\frac{a^{3/2} (c-11 d) (c+d)^2 \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{8 d^{3/2} f}+\frac{a^2 (c-11 d) (c+d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{8 d f \sqrt{a+a \sin (e+f x)}}+\frac{a^2 (c-11 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{12 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{3 d f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.834184, size = 281, normalized size = 1.23 \[ \frac{(a (\sin (e+f x)+1))^{3/2} \left (\frac{(c-11 d) (c+d)^2 \left (-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )+\log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )\right )}{d^{3/2}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sqrt{c+d \sin (e+f x)} \left (3 c^2+2 d (7 c+11 d) \sin (e+f x)+52 c d-4 d^2 \cos (2 (e+f x))+37 d^2\right )}{3 d}\right )}{16 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*(((c - 11*d)*(c + d)^2*(-2*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sq
rt[c + d*Sin[e + f*x]]] - ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + Log[
Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(3/2) - (2*(Cos[(e + f*x)/2] - Sin[(
e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(3*c^2 + 52*c*d + 37*d^2 - 4*d^2*Cos[2*(e + f*x)] + 2*d*(7*c + 11*d)*Sin
[e + f*x]))/(3*d)))/(16*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \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))^(3/2),x)

[Out]

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

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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 )}^{\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)*(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 9.25408, size = 3214, normalized size = 14.1 \begin{align*} \text{result too large to display} \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))^(3/2),x, algorithm="fricas")

[Out]

[-1/192*(3*(a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3 + (a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3)*cos(f*x +
e) + (a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3)*sin(f*x + e))*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^
4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a*c^2*d^2 - 1
4*a*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 32*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16
*d^4*cos(f*x + e)^4 - c^3*d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*(5*c^2*d^2
- 26*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3*d - 7*c^2*d^2 + 31*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x +
e)^3 + c^3*d - 17*c^2*d^2 + 59*c*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 14*c*d^3
+ 13*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 -
28*a*c^3*d + 230*a*c^2*d^2 - 476*a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c
^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 +
5*a*d^4)*cos(f*x + e)^2 + 32*(a*c^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(cos(f
*x + e) + sin(f*x + e) + 1)) - 8*(8*a*d^2*cos(f*x + e)^3 - 3*a*c^2 - 38*a*c*d - 19*a*d^2 - 14*(a*c*d + a*d^2)*
cos(f*x + e)^2 - (3*a*c^2 + 52*a*c*d + 41*a*d^2)*cos(f*x + e) - (8*a*d^2*cos(f*x + e)^2 - 3*a*c^2 - 38*a*c*d -
19*a*d^2 + 2*(7*a*c*d + 11*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) +
c))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f), -1/96*(3*(a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3 + (a*c^3
- 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3)*cos(f*x + e) + (a*c^3 - 9*a*c^2*d - 21*a*c*d^2 - 11*a*d^3)*sin(f*x + e))*
sqrt(a/d)*arctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2 + 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x
+ e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(a/d)/(2*a*d^2*cos(f*x + e)^3 - (3*a*c*d - a*d^2)*cos(f*x + e)*sin(f*x
+ e) - (a*c^2 - a*c*d + 2*a*d^2)*cos(f*x + e))) - 4*(8*a*d^2*cos(f*x + e)^3 - 3*a*c^2 - 38*a*c*d - 19*a*d^2 -
14*(a*c*d + a*d^2)*cos(f*x + e)^2 - (3*a*c^2 + 52*a*c*d + 41*a*d^2)*cos(f*x + e) - (8*a*d^2*cos(f*x + e)^2 - 3
*a*c^2 - 38*a*c*d - 19*a*d^2 + 2*(7*a*c*d + 11*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqr
t(d*sin(f*x + e) + c))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f)]

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Sympy [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))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError