3.81 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=198 \[ \frac{256 a c^5 (11 A-5 B) \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a c^4 (11 A-5 B) \cos ^3(e+f x)}{1155 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 (11 A-5 B) \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{231 f}+\frac{2 a c^2 (11 A-5 B) \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

[Out]

(256*a*(11*A - 5*B)*c^5*Cos[e + f*x]^3)/(3465*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*(11*A - 5*B)*c^4*Cos[e + f
*x]^3)/(1155*f*Sqrt[c - c*Sin[e + f*x]]) + (8*a*(11*A - 5*B)*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(231
*f) + (2*a*(11*A - 5*B)*c^2*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(99*f) - (2*a*B*c*Cos[e + f*x]^3*(c - c
*Sin[e + f*x])^(5/2))/(11*f)

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Rubi [A]  time = 0.487202, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2967, 2856, 2674, 2673} \[ \frac{256 a c^5 (11 A-5 B) \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a c^4 (11 A-5 B) \cos ^3(e+f x)}{1155 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 (11 A-5 B) \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{231 f}+\frac{2 a c^2 (11 A-5 B) \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(256*a*(11*A - 5*B)*c^5*Cos[e + f*x]^3)/(3465*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*(11*A - 5*B)*c^4*Cos[e + f
*x]^3)/(1155*f*Sqrt[c - c*Sin[e + f*x]]) + (8*a*(11*A - 5*B)*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(231
*f) + (2*a*(11*A - 5*B)*c^2*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(99*f) - (2*a*B*c*Cos[e + f*x]^3*(c - c
*Sin[e + f*x])^(5/2))/(11*f)

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2856

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac{1}{11} (a (11 A-5 B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx\\ &=\frac{2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac{1}{33} \left (4 a (11 A-5 B) c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac{8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{231 f}+\frac{2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac{1}{231} \left (32 a (11 A-5 B) c^3\right ) \int \cos ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{231 f}+\frac{2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}+\frac{\left (128 a (11 A-5 B) c^4\right ) \int \frac{\cos ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{1155}\\ &=\frac{256 a (11 A-5 B) c^5 \cos ^3(e+f x)}{3465 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a (11 A-5 B) c^4 \cos ^3(e+f x)}{1155 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a (11 A-5 B) c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{231 f}+\frac{2 a (11 A-5 B) c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{99 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{11 f}\\ \end{align*}

Mathematica [A]  time = 2.88061, size = 149, normalized size = 0.75 \[ -\frac{a c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 (60 (121 A-202 B) \cos (2 (e+f x))+30558 A \sin (e+f x)-770 A \sin (3 (e+f x))-35332 A-31530 B \sin (e+f x)+2870 B \sin (3 (e+f x))+315 B \cos (4 (e+f x))+27085 B)}{13860 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a*c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-35332*A + 27085*B + 60*(121*A - 202
*B)*Cos[2*(e + f*x)] + 315*B*Cos[4*(e + f*x)] + 30558*A*Sin[e + f*x] - 31530*B*Sin[e + f*x] - 770*A*Sin[3*(e +
f*x)] + 2870*B*Sin[3*(e + f*x)]))/(13860*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]  time = 1.046, size = 119, normalized size = 0.6 \begin{align*}{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{4} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}a \left ( \left ( -385\,A+1435\,B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( 3916\,A-4300\,B \right ) \sin \left ( fx+e \right ) +315\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( 1815\,A-3345\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-5324\,A+4940\,B \right ) }{3465\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{c-c\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))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x)

[Out]

2/3465*(-1+sin(f*x+e))*c^4*(1+sin(f*x+e))^2*a*((-385*A+1435*B)*sin(f*x+e)*cos(f*x+e)^2+(3916*A-4300*B)*sin(f*x
+e)+315*B*cos(f*x+e)^4+(1815*A-3345*B)*cos(f*x+e)^2-5324*A+4940*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(7/2), x)

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Fricas [A]  time = 1.78942, size = 749, normalized size = 3.78 \begin{align*} \frac{2 \,{\left (315 \, B a c^{3} \cos \left (f x + e\right )^{6} - 35 \,{\left (11 \, A - 32 \, B\right )} a c^{3} \cos \left (f x + e\right )^{5} + 5 \,{\left (209 \, A - 221 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 2 \,{\left (1243 \, A - 1195 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 32 \,{\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} + 128 \,{\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) + 256 \,{\left (11 \, A - 5 \, B\right )} a c^{3} -{\left (315 \, B a c^{3} \cos \left (f x + e\right )^{5} + 35 \,{\left (11 \, A - 23 \, B\right )} a c^{3} \cos \left (f x + e\right )^{4} + 10 \,{\left (143 \, A - 191 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - 96 \,{\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right )^{2} - 128 \,{\left (11 \, A - 5 \, B\right )} a c^{3} \cos \left (f x + e\right ) - 256 \,{\left (11 \, A - 5 \, B\right )} a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3465 \,{\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))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/3465*(315*B*a*c^3*cos(f*x + e)^6 - 35*(11*A - 32*B)*a*c^3*cos(f*x + e)^5 + 5*(209*A - 221*B)*a*c^3*cos(f*x +
e)^4 + 2*(1243*A - 1195*B)*a*c^3*cos(f*x + e)^3 - 32*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 + 128*(11*A - 5*B)*a*c
^3*cos(f*x + e) + 256*(11*A - 5*B)*a*c^3 - (315*B*a*c^3*cos(f*x + e)^5 + 35*(11*A - 23*B)*a*c^3*cos(f*x + e)^4
+ 10*(143*A - 191*B)*a*c^3*cos(f*x + e)^3 - 96*(11*A - 5*B)*a*c^3*cos(f*x + e)^2 - 128*(11*A - 5*B)*a*c^3*cos
(f*x + e) - 256*(11*A - 5*B)*a*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) +
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))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(7/2), x)