3.8.24 \(\int (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx\) [724]

Optimal. Leaf size=235 \[ -\frac {2 (3 b c+5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac {2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (3 b c+5 a d) \left (c^2-d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/5*b*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f-2/15*(5*a*d+3*b*c)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f-2/15*(20*a*c
*d+3*b*(c^2+3*d^2))*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1
/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/d/f/((c+d*sin(f*x+e))/(c+d))^(1/2)+2/15*(5*a*d+3*b*c)*
(c^2-d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^
(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/d/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2832, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {2 \left (c^2-d^2\right ) (5 a d+3 b c) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (5 a d+3 b c) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*b*c + 5*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*f) - (2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/
2))/(5*f) + (2*(20*a*c*d + 3*b*(c^2 + 3*d^2))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e +
f*x]])/(15*d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*(3*b*c + 5*a*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)
/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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

Rule 2831

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

Rule 2832

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[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx &=-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac {2}{5} \int \sqrt {c+d \sin (e+f x)} \left (\frac {1}{2} (5 a c+3 b d)+\frac {1}{2} (3 b c+5 a d) \sin (e+f x)\right ) \, dx\\ &=-\frac {2 (3 b c+5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac {4}{15} \int \frac {\frac {1}{4} \left (12 b c d+5 a \left (3 c^2+d^2\right )\right )+\frac {1}{4} \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx\\ &=-\frac {2 (3 b c+5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}-\frac {\left ((3 b c+5 a d) \left (c^2-d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d}+\frac {\left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 d}\\ &=-\frac {2 (3 b c+5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac {\left (\left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{15 d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {\left ((3 b c+5 a d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{15 d \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {2 (3 b c+5 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 f}+\frac {2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (3 b c+5 a d) \left (c^2-d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.81, size = 218, normalized size = 0.93 \begin {gather*} \frac {-2 d \left (12 b c d+5 a \left (3 c^2+d^2\right )\right ) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-2 \left (20 a c d+3 b \left (c^2+3 d^2\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-2 d \cos (e+f x) (c+d \sin (e+f x)) (6 b c+5 a d+3 b d \sin (e+f x))}{15 d f \sqrt {c+d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(12*b*c*d + 5*a*(3*c^2 + d^2))*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])
/(c + d)] - 2*(20*a*c*d + 3*b*(c^2 + 3*d^2))*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*Elli
pticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - 2*d*Cos[e + f*x]*(c + d*Sin[
e + f*x])*(6*b*c + 5*a*d + 3*b*d*Sin[e + f*x]))/(15*d*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1448\) vs. \(2(281)=562\).
time = 7.01, size = 1449, normalized size = 6.17

method	result	size
default	\(\text {Expression too large to display}\)	\(1449\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*a*c^3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)
*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d+5*a*c^2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+
sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+
d))^(1/2))*d^2-15*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1
/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c*d^3-5*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(
-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/
(c+d))^(1/2))*a*d^4+3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d)
)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^3*d+9*((c+d*sin(f*x+e))/(c-d))^(1/2)
*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c
-d)/(c+d))^(1/2))*b*c^2*d^2-3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e
))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c*d^3-9*((c+d*sin(f*x+e))/(c-d
))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(
1/2),((c-d)/(c+d))^(1/2))*b*d^4-20*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(
f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c^3*d+20*((c+d*sin(f*x+e)
)/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c
-d))^(1/2),((c-d)/(c+d))^(1/2))*a*c*d^3-3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*
(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^4-6*((c+d*sin(f*
x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e)
)/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*c^2*d^2+9*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(-1+sin(f*x+e))*d/(c+d))^(1/2
)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*b*d^4+3*b*d^4*
sin(f*x+e)^4+5*a*d^4*sin(f*x+e)^3+9*b*c*d^3*sin(f*x+e)^3+5*a*c*d^3*sin(f*x+e)^2+6*b*c^2*d^2*sin(f*x+e)^2-3*b*d
^4*sin(f*x+e)^2-5*a*d^4*sin(f*x+e)-9*b*c*d^3*sin(f*x+e)-5*a*c*d^3-6*b*c^2*d^2)/d^2/cos(f*x+e)/(c+d*sin(f*x+e))
^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.16, size = 521, normalized size = 2.22 \begin {gather*} -\frac {\sqrt {2} {\left (6 \, b c^{3} - 5 \, a c^{2} d - 18 \, b c d^{2} - 15 \, a d^{3}\right )} \sqrt {i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right ) + \sqrt {2} {\left (6 \, b c^{3} - 5 \, a c^{2} d - 18 \, b c d^{2} - 15 \, a d^{3}\right )} \sqrt {-i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right ) + 3 \, \sqrt {2} {\left (3 i \, b c^{2} d + 20 i \, a c d^{2} + 9 i \, b d^{3}\right )} \sqrt {i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, b c^{2} d - 20 i \, a c d^{2} - 9 i \, b d^{3}\right )} \sqrt {-i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right )\right ) + 6 \, {\left (3 \, b d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (6 \, b c d^{2} + 5 \, a d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{45 \, d^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/45*(sqrt(2)*(6*b*c^3 - 5*a*c^2*d - 18*b*c*d^2 - 15*a*d^3)*sqrt(I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2
)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) + sqrt(2)*(6*b*
c^3 - 5*a*c^2*d - 18*b*c*d^2 - 15*a*d^3)*sqrt(-I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*
c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) + 3*sqrt(2)*(3*I*b*c^2*d + 20*I*a
*c*d^2 + 9*I*b*d^3)*sqrt(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weier
strassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*
x + e) - 2*I*c)/d)) + 3*sqrt(2)*(-3*I*b*c^2*d - 20*I*a*c*d^2 - 9*I*b*d^3)*sqrt(-I*d)*weierstrassZeta(-4/3*(4*c
^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c
^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)) + 6*(3*b*d^3*cos(f*x + e)*sin(f*x
+ e) + (6*b*c*d^2 + 5*a*d^3)*cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(d^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (a+b\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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