3.163 \(\int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=142 \[ \frac{c^2 (3 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{30 f \sqrt{c-c \sin (e+f x)}}+\frac{c (3 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt{c-c \sin (e+f x)}}{15 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f} \]

[Out]

((3*A + B)*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(30*f*Sqrt[c - c*Sin[e + f*x]]) + ((3*A + B)*c*Cos[e +
f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(15*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)
*(c - c*Sin[e + f*x])^(3/2))/(6*f)

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Rubi [A]  time = 0.359294, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {2973, 2740, 2738} \[ \frac{c^2 (3 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{30 f \sqrt{c-c \sin (e+f x)}}+\frac{c (3 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt{c-c \sin (e+f x)}}{15 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

((3*A + B)*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(30*f*Sqrt[c - c*Sin[e + f*x]]) + ((3*A + B)*c*Cos[e +
f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(15*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)
*(c - c*Sin[e + f*x])^(3/2))/(6*f)

Rule 2973

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

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2738

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] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}+\frac{1}{3} (3 A+B) \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac{(3 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)}}{15 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}+\frac{1}{15} (2 (3 A+B) c) \int (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{(3 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{30 f \sqrt{c-c \sin (e+f x)}}+\frac{(3 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)}}{15 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}\\ \end{align*}

Mathematica [A]  time = 1.85639, size = 212, normalized size = 1.49 \[ -\frac{a^3 c (\sin (e+f x)-1) (\sin (e+f x)+1)^3 \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (-15 (16 A+11 B) \cos (2 (e+f x))-30 (2 A+B) \cos (4 (e+f x))+840 A \sin (e+f x)+60 A \sin (3 (e+f x))-12 A \sin (5 (e+f x))+240 B \sin (e+f x)-40 B \sin (3 (e+f x))-24 B \sin (5 (e+f x))+5 B \cos (6 (e+f x)))}{960 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-(a^3*c*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-15*(16*
A + 11*B)*Cos[2*(e + f*x)] - 30*(2*A + B)*Cos[4*(e + f*x)] + 5*B*Cos[6*(e + f*x)] + 840*A*Sin[e + f*x] + 240*B
*Sin[e + f*x] + 60*A*Sin[3*(e + f*x)] - 40*B*Sin[3*(e + f*x)] - 12*A*Sin[5*(e + f*x)] - 24*B*Sin[5*(e + f*x)])
)/(960*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7)

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Maple [A]  time = 0.329, size = 185, normalized size = 1.3 \begin{align*}{\frac{ \left ( 5\,B\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+6\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+12\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}-15\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -10\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -12\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-15\,A\sin \left ( fx+e \right ) -10\,B\sin \left ( fx+e \right ) -24\,A-8\,B \right ) \sin \left ( fx+e \right ) }{30\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) -2 \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x)

[Out]

1/30/f*(5*B*sin(f*x+e)*cos(f*x+e)^4+6*A*cos(f*x+e)^4+12*B*cos(f*x+e)^4-15*A*cos(f*x+e)^2*sin(f*x+e)-10*B*cos(f
*x+e)^2*sin(f*x+e)-12*A*cos(f*x+e)^2-4*B*cos(f*x+e)^2-15*A*sin(f*x+e)-10*B*sin(f*x+e)-24*A-8*B)*(-c*(-1+sin(f*
x+e)))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)/(cos(f*x+e)^2-2*sin(f*x+e)-2)/cos(f*x+e)^3

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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 )}^{\frac{7}{2}}{\left (-c \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))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(3/2), x)

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Fricas [A]  time = 1.56558, size = 356, normalized size = 2.51 \begin{align*} \frac{{\left (5 \, B a^{3} c \cos \left (f x + e\right )^{6} - 15 \,{\left (A + B\right )} a^{3} c \cos \left (f x + e\right )^{4} + 5 \,{\left (3 \, A + 2 \, B\right )} a^{3} c - 2 \,{\left (3 \,{\left (A + 2 \, B\right )} a^{3} c \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, A + B\right )} a^{3} c \cos \left (f x + e\right )^{2} - 4 \,{\left (3 \, A + B\right )} a^{3} c\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/30*(5*B*a^3*c*cos(f*x + e)^6 - 15*(A + B)*a^3*c*cos(f*x + e)^4 + 5*(3*A + 2*B)*a^3*c - 2*(3*(A + 2*B)*a^3*c*
cos(f*x + e)^4 - 2*(3*A + B)*a^3*c*cos(f*x + e)^2 - 4*(3*A + B)*a^3*c)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*
sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

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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))**(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [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))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out