3.6.37 \(\int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))^2 \, dx\) [537]

Optimal. Leaf size=202 \[ -\frac {64 a^3 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x)}{315 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}-\frac {2 a \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac {4 (9 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f} \]

[Out]

-2/105*a*(21*c^2+30*c*d+13*d^2)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-4/63*(9*c-d)*d*cos(f*x+e)*(a+a*sin(f*x+e))
^(5/2)/f-2/9*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f-64/315*a^3*(21*c^2+30*c*d+13*d^2)*cos(f*x+e)/f/(a+a*sin
(f*x+e))^(1/2)-16/315*a^2*(21*c^2+30*c*d+13*d^2)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.19, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2840, 2830, 2726, 2725} \begin {gather*} -\frac {64 a^3 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x)}{315 f \sqrt {a \sin (e+f x)+a}}-\frac {16 a^2 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{315 f}-\frac {2 a \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 f}-\frac {4 d (9 c-d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{63 f}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*(21*c^2 + 30*c*d + 13*d^2)*Cos[e + f*x])/(315*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*(21*c^2 + 30*c*d
+ 13*d^2)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(315*f) - (2*a*(21*c^2 + 30*c*d + 13*d^2)*Cos[e + f*x]*(a + a
*Sin[e + f*x])^(3/2))/(105*f) - (4*(9*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(63*f) - (2*d^2*Cos[e
+ f*x]*(a + a*Sin[e + f*x])^(7/2))/(9*a*f)

Rule 2725

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]

Rule 2726

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] &&
EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2830

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*S
in[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 2840

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]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))^2 \, dx &=-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac {2 \int (a+a \sin (e+f x))^{5/2} \left (\frac {1}{2} a \left (9 c^2+7 d^2\right )+a (9 c-d) d \sin (e+f x)\right ) \, dx}{9 a}\\ &=-\frac {4 (9 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac {1}{21} \left (21 c^2+30 c d+13 d^2\right ) \int (a+a \sin (e+f x))^{5/2} \, dx\\ &=-\frac {2 a \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac {4 (9 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac {1}{105} \left (8 a \left (21 c^2+30 c d+13 d^2\right )\right ) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac {16 a^2 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}-\frac {2 a \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac {4 (9 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac {1}{315} \left (32 a^2 \left (21 c^2+30 c d+13 d^2\right )\right ) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=-\frac {64 a^3 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x)}{315 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}-\frac {2 a \left (21 c^2+30 c d+13 d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac {4 (9 c-d) d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac {2 d^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}\\ \end {align*}

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Mathematica [A]
time = 2.11, size = 180, normalized size = 0.89 \begin {gather*} -\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (7476 c^2+12480 c d+5653 d^2-4 \left (63 c^2+360 c d+254 d^2\right ) \cos (2 (e+f x))+35 d^2 \cos (4 (e+f x))+2352 c^2 \sin (e+f x)+6060 c d \sin (e+f x)+3116 d^2 \sin (e+f x)-180 c d \sin (3 (e+f x))-260 d^2 \sin (3 (e+f x))\right )}{1260 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1260*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(7476*c^2 + 12480*c*d + 5653*d^2
- 4*(63*c^2 + 360*c*d + 254*d^2)*Cos[2*(e + f*x)] + 35*d^2*Cos[4*(e + f*x)] + 2352*c^2*Sin[e + f*x] + 6060*c*
d*Sin[e + f*x] + 3116*d^2*Sin[e + f*x] - 180*c*d*Sin[3*(e + f*x)] - 260*d^2*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2]))

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Maple [A]
time = 3.09, size = 168, normalized size = 0.83

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (35 d^{2} \left (\sin ^{4}\left (f x +e \right )\right )+90 c d \left (\sin ^{3}\left (f x +e \right )\right )+130 d^{2} \left (\sin ^{3}\left (f x +e \right )\right )+63 c^{2} \left (\sin ^{2}\left (f x +e \right )\right )+360 c d \left (\sin ^{2}\left (f x +e \right )\right )+219 d^{2} \left (\sin ^{2}\left (f x +e \right )\right )+294 c^{2} \sin \left (f x +e \right )+690 c d \sin \left (f x +e \right )+292 d^{2} \sin \left (f x +e \right )+903 c^{2}+1380 c d +584 d^{2}\right )}{315 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(1+sin(f*x+e))*a^3*(sin(f*x+e)-1)*(35*d^2*sin(f*x+e)^4+90*c*d*sin(f*x+e)^3+130*d^2*sin(f*x+e)^3+63*c^2*s
in(f*x+e)^2+360*c*d*sin(f*x+e)^2+219*d^2*sin(f*x+e)^2+294*c^2*sin(f*x+e)+690*c*d*sin(f*x+e)+292*d^2*sin(f*x+e)
+903*c^2+1380*c*d+584*d^2)/cos(f*x+e)/(a+a*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+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 352, normalized size = 1.74 \begin {gather*} -\frac {2 \, {\left (35 \, a^{2} d^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (18 \, a^{2} c d + 19 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{4} + 672 \, a^{2} c^{2} + 960 \, a^{2} c d + 416 \, a^{2} d^{2} - {\left (63 \, a^{2} c^{2} + 360 \, a^{2} c d + 289 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (231 \, a^{2} c^{2} + 510 \, a^{2} c d + 263 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (483 \, a^{2} c^{2} + 870 \, a^{2} c d + 419 \, a^{2} d^{2}\right )} \cos \left (f x + e\right ) - {\left (35 \, a^{2} d^{2} \cos \left (f x + e\right )^{4} + 672 \, a^{2} c^{2} + 960 \, a^{2} c d + 416 \, a^{2} d^{2} + 10 \, {\left (9 \, a^{2} c d + 13 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (21 \, a^{2} c^{2} + 90 \, a^{2} c d + 53 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (147 \, a^{2} c^{2} + 390 \, a^{2} c d + 211 \, a^{2} 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}}{315 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*a^2*d^2*cos(f*x + e)^5 - 5*(18*a^2*c*d + 19*a^2*d^2)*cos(f*x + e)^4 + 672*a^2*c^2 + 960*a^2*c*d + 4
16*a^2*d^2 - (63*a^2*c^2 + 360*a^2*c*d + 289*a^2*d^2)*cos(f*x + e)^3 + (231*a^2*c^2 + 510*a^2*c*d + 263*a^2*d^
2)*cos(f*x + e)^2 + 2*(483*a^2*c^2 + 870*a^2*c*d + 419*a^2*d^2)*cos(f*x + e) - (35*a^2*d^2*cos(f*x + e)^4 + 67
2*a^2*c^2 + 960*a^2*c*d + 416*a^2*d^2 + 10*(9*a^2*c*d + 13*a^2*d^2)*cos(f*x + e)^3 - 3*(21*a^2*c^2 + 90*a^2*c*
d + 53*a^2*d^2)*cos(f*x + e)^2 - 2*(147*a^2*c^2 + 390*a^2*c*d + 211*a^2*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**(5/2)*(c + d*sin(e + f*x))**2, x)

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Giac [A]
time = 0.63, size = 348, normalized size = 1.72 \begin {gather*} \frac {\sqrt {2} {\left (35 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) + 630 \, {\left (20 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 30 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 13 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 420 \, {\left (5 \, a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 11 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 252 \, {\left (a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 45 \, {\left (4 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right )\right )} \sqrt {a}}{2520 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*sqrt(2)*(35*a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-9/4*pi + 9/2*f*x + 9/2*e) + 630*(20*a^2*c^
2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 30*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 13*a^2*d^2*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 420*(5*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
+ 11*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi
+ 3/2*f*x + 3/2*e) + 252*(a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x +
1/2*e)) + 3*a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-5/4*pi + 5/2*f*x + 5/2*e) + 45*(4*a^2*c*d*sgn(c
os(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-7/4*pi + 7/2*f*x + 7/2*e)
)*sqrt(a)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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