3.489 \(\int (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=378 \[ \frac {4 a^2 \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d f}+\frac {4 a^2 \left (c^2-d^2\right ) \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \left (5 c^4-45 c^3 d-381 c^2 d^2-435 c d^3-168 d^4\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f} \]

[Out]

4/315*a^2*(5*c*(c-9*d)-56*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d/f+4/63*a^2*(c-9*d)*cos(f*x+e)*(c+d*sin(f*x+
e))^(5/2)/d/f-2/9*a^2*cos(f*x+e)*(c+d*sin(f*x+e))^(7/2)/d/f+4/315*a^2*(5*c^3-45*c^2*d-141*c*d^2-75*d^3)*cos(f*
x+e)*(c+d*sin(f*x+e))^(1/2)/d/f+4/315*a^2*(5*c^4-45*c^3*d-381*c^2*d^2-435*c*d^3-168*d^4)*(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^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-4/315*a^2*(c^2-d^2)*(5*c^3-45*c^2*d-141*c*d^2-75*d^3)*(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^2/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.67, antiderivative size = 378, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2763, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {4 a^2 \left (-45 c^2 d+5 c^3-141 c d^2-75 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d f}+\frac {4 a^2 \left (c^2-d^2\right ) \left (-45 c^2 d+5 c^3-141 c d^2-75 d^3\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \left (-381 c^2 d^2-45 c^3 d+5 c^4-435 c d^3-168 d^4\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(4*a^2*(5*c^3 - 45*c^2*d - 141*c*d^2 - 75*d^3)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(315*d*f) + (4*a^2*(5*c*
(c - 9*d) - 56*d^2)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(315*d*f) + (4*a^2*(c - 9*d)*Cos[e + f*x]*(c + d*
Sin[e + f*x])^(5/2))/(63*d*f) - (2*a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(7/2))/(9*d*f) - (4*a^2*(5*c^4 - 45*c
^3*d - 381*c^2*d^2 - 435*c*d^3 - 168*d^4)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]
])/(315*d^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c^2 - d^2)*(5*c^3 - 45*c^2*d - 141*c*d^2 - 75*d^3)
*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(315*d^2*f*Sqrt[c + d*Sin[e
+ f*x]])

Rule 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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 2753

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)*Simp[
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]

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]))

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{5/2} \, dx &=-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {2 \int \left (8 a^2 d-a^2 (c-9 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{5/2} \, dx}{9 d}\\ &=\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {4 \int (c+d \sin (e+f x))^{3/2} \left (\frac {3}{2} a^2 d (17 c+15 d)-\frac {1}{2} a^2 \left (5 c (c-9 d)-56 d^2\right ) \sin (e+f x)\right ) \, dx}{63 d}\\ &=\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {8 \int \sqrt {c+d \sin (e+f x)} \left (6 a^2 d \left (10 c^2+15 c d+7 d^2\right )-\frac {3}{4} a^2 \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \sin (e+f x)\right ) \, dx}{315 d}\\ &=\frac {4 a^2 \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d f}+\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {16 \int \frac {\frac {3}{8} a^2 d \left (235 c^3+405 c^2 d+309 c d^2+75 d^3\right )-\frac {3}{8} a^2 \left (5 c^4-45 c^3 d-381 c^2 d^2-435 c d^3-168 d^4\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{945 d}\\ &=\frac {4 a^2 \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d f}+\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}+\frac {\left (2 a^2 \left (c^2-d^2\right ) \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{315 d^2}-\frac {\left (2 a^2 \left (5 c^4-45 c^3 d-381 c^2 d^2-435 c d^3-168 d^4\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{315 d^2}\\ &=\frac {4 a^2 \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d f}+\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}-\frac {\left (2 a^2 \left (5 c^4-45 c^3 d-381 c^2 d^2-435 c d^3-168 d^4\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{315 d^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^2 \left (c^2-d^2\right ) \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\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}{315 d^2 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {4 a^2 \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d f}+\frac {4 a^2 \left (5 c (c-9 d)-56 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d f}+\frac {4 a^2 (c-9 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d f}-\frac {2 a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{9 d f}-\frac {4 a^2 \left (5 c^4-45 c^3 d-381 c^2 d^2-435 c d^3-168 d^4\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)}}{315 d^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 \left (c^2-d^2\right ) \left (5 c^3-45 c^2 d-141 c d^2-75 d^3\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}}}{315 d^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 1.98, size = 322, normalized size = 0.85 \[ \frac {a^2 (\sin (e+f x)+1)^2 \left (16 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left (\left (5 c^4-45 c^3 d-381 c^2 d^2-435 c d^3-168 d^4\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )-d^2 \left (235 c^3+405 c^2 d+309 c d^2+75 d^3\right ) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )-d (c+d \sin (e+f x)) \left (2 d \left (\sin (2 (e+f x)) \left (150 c^2+540 c d-35 d^2 \cos (2 (e+f x))+259 d^2\right )-5 d (19 c+18 d) \cos (3 (e+f x))\right )+2 \left (20 c^3+1080 c^2 d+1671 c d^2+690 d^3\right ) \cos (e+f x)\right )\right )}{1260 d^2 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(a^2*(1 + Sin[e + f*x])^2*(16*(-(d^2*(235*c^3 + 405*c^2*d + 309*c*d^2 + 75*d^3)*EllipticF[(-2*e + Pi - 2*f*x)/
4, (2*d)/(c + d)]) + (5*c^4 - 45*c^3*d - 381*c^2*d^2 - 435*c*d^3 - 168*d^4)*((c + d)*EllipticE[(-2*e + Pi - 2*
f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]))*Sqrt[(c + d*Sin[e + f*x])/(c + d)
] - d*(c + d*Sin[e + f*x])*(2*(20*c^3 + 1080*c^2*d + 1671*c*d^2 + 690*d^3)*Cos[e + f*x] + 2*d*(-5*d*(19*c + 18
*d)*Cos[3*(e + f*x)] + (150*c^2 + 540*c*d + 259*d^2 - 35*d^2*Cos[2*(e + f*x)])*Sin[2*(e + f*x)]))))/(1260*d^2*
f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c + d*Sin[e + f*x]])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} d^{2} \cos \left (f x + e\right )^{4} + 2 \, a^{2} c^{2} + 4 \, a^{2} c d + 2 \, a^{2} d^{2} - {\left (a^{2} c^{2} + 4 \, a^{2} c d + 3 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} c^{2} + 2 \, a^{2} c d + a^{2} d^{2} - {\left (a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral((a^2*d^2*cos(f*x + e)^4 + 2*a^2*c^2 + 4*a^2*c*d + 2*a^2*d^2 - (a^2*c^2 + 4*a^2*c*d + 3*a^2*d^2)*cos(f
*x + e)^2 + 2*(a^2*c^2 + 2*a^2*c*d + a^2*d^2 - (a^2*c*d + a^2*d^2)*cos(f*x + e)^2)*sin(f*x + e))*sqrt(d*sin(f*
x + e) + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^(5/2), x)

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maple [B]  time = 1.59, size = 1614, normalized size = 4.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^(5/2),x)

[Out]

-2/315*a^2*(-426*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^2*d^4+5*c^4*d^2+90*((c+d*sin(f*x+e))/(c-d))^
(1/2)*(-(sin(f*x+e)-1)*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))*c^5*d-870*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c*d^5+10*((c+d*sin(f*x+e))/(c-d
))^(1/2)*(-(sin(f*x+e)-1)*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))*c^5*d-570*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^4*d^2-1012*((c+d*sin(f*x+e
))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^3*d^3+84*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^2*d^4+1002*((c+d*s
in(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c*d^5+772*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^4*d^2+780*(
(c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*c^3*d^3+270*c^3*d^3+150*c*d^5-10*((c+d*sin(f*x+e))/(c-d))^(1/2)*
(-(sin(f*x+e)-1)*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))*c^6-336*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*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))*d^6+486*((c+d*sin(f*x+e))/(c-d))^(1/2)*
(-(sin(f*x+e)-1)*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^6-130*c*d^5*sin(f*x+e)^5-170*c^2*d^4*sin(f*x+e)^4-360*c*d^5*sin(f*x+e)^4-80*c^3*d^3*sin(f*x+
e)^3-540*c^2*d^4*sin(f*x+e)^3-376*c*d^5*sin(f*x+e)^3-5*c^4*d^2*sin(f*x+e)^2-270*c^3*d^3*sin(f*x+e)^2-224*c^2*d
^4*sin(f*x+e)^2+210*c*d^5*sin(f*x+e)^2+80*c^3*d^3*sin(f*x+e)+540*c^2*d^4*sin(f*x+e)+506*c*d^5*sin(f*x+e)-90*d^
6*sin(f*x+e)^5-77*d^6*sin(f*x+e)^4-60*d^6*sin(f*x+e)^3+112*d^6*sin(f*x+e)^2+150*d^6*sin(f*x+e)-35*d^6*sin(f*x+
e)^6+394*c^2*d^4)/d^3/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^(5/2),x)

[Out]

int((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int c^{2} \sqrt {c + d \sin {\left (e + f x \right )}}\, dx + \int 2 c^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int c^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int 2 d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )}\, dx + \int d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{4}{\left (e + f x \right )}\, dx + \int 2 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int 4 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int 2 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2*(c+d*sin(f*x+e))**(5/2),x)

[Out]

a**2*(Integral(c**2*sqrt(c + d*sin(e + f*x)), x) + Integral(2*c**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) +
Integral(c**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(d**2*sqrt(c + d*sin(e + f*x))*sin(e + f
*x)**2, x) + Integral(2*d**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3, x) + Integral(d**2*sqrt(c + d*sin(e + f
*x))*sin(e + f*x)**4, x) + Integral(2*c*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integral(4*c*d*sqrt(c +
d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(2*c*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3, x))

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