3.162 \(\int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=192 \[ \frac{2 c^2 (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt{c-c \sin (e+f x)}}{105 f}+\frac{c^3 (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{c (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f} \]

[Out]

((7*A + B)*c^3*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*(7*A + B)*c^2*Co
s[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(105*f) + ((7*A + B)*c*Cos[e + f*x]*(a + a*Sin
[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2))/(42*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e
+ f*x])^(5/2))/(7*f)

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Rubi [A]  time = 0.459797, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {2973, 2740, 2738} \[ \frac{2 c^2 (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt{c-c \sin (e+f x)}}{105 f}+\frac{c^3 (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{c (7 A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f} \]

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])^(5/2),x]

[Out]

((7*A + B)*c^3*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*(7*A + B)*c^2*Co
s[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(105*f) + ((7*A + B)*c*Cos[e + f*x]*(a + a*Sin
[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2))/(42*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e
+ f*x])^(5/2))/(7*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))^{5/2} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}+\frac{1}{7} (7 A+B) \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx\\ &=\frac{(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}+\frac{1}{21} (2 (7 A+B) c) \int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac{2 (7 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)}}{105 f}+\frac{(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}+\frac{1}{105} \left (4 (7 A+B) c^2\right ) \int (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{(7 A+B) c^3 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{2 (7 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)}}{105 f}+\frac{(7 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{42 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2}}{7 f}\\ \end{align*}

Mathematica [A]  time = 2.19684, size = 232, normalized size = 1.21 \[ \frac{a^3 c^2 (\sin (e+f x)-1)^2 (\sin (e+f x)+1)^3 \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (-525 (A+B) \cos (2 (e+f x))-210 (A+B) \cos (4 (e+f x))+4200 A \sin (e+f x)+700 A \sin (3 (e+f x))+84 A \sin (5 (e+f x))-35 A \cos (6 (e+f x))+525 B \sin (e+f x)-35 B \sin (3 (e+f x))-63 B \sin (5 (e+f x))-15 B \sin (7 (e+f x))-35 B \cos (6 (e+f x)))}{6720 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

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])^(5/2),x]

[Out]

(a^3*c^2*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-525*
(A + B)*Cos[2*(e + f*x)] - 210*(A + B)*Cos[4*(e + f*x)] - 35*A*Cos[6*(e + f*x)] - 35*B*Cos[6*(e + f*x)] + 4200
*A*Sin[e + f*x] + 525*B*Sin[e + f*x] + 700*A*Sin[3*(e + f*x)] - 35*B*Sin[3*(e + f*x)] + 84*A*Sin[5*(e + f*x)]
- 63*B*Sin[5*(e + f*x)] - 15*B*Sin[7*(e + f*x)]))/(6720*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^7)

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Maple [A]  time = 0.359, size = 203, normalized size = 1.1 \begin{align*}{\frac{ \left ( -30\,B \left ( \cos \left ( fx+e \right ) \right ) ^{6}+35\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}\sin \left ( fx+e \right ) +35\,B\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+42\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+6\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+35\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +35\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +56\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+35\,A\sin \left ( fx+e \right ) +35\,B\sin \left ( fx+e \right ) +112\,A+16\,B \right ) \sin \left ( fx+e \right ) }{210\,f \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{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))^(5/2),x)

[Out]

1/210/f*(-30*B*cos(f*x+e)^6+35*A*cos(f*x+e)^4*sin(f*x+e)+35*B*sin(f*x+e)*cos(f*x+e)^4+42*A*cos(f*x+e)^4+6*B*co
s(f*x+e)^4+35*A*cos(f*x+e)^2*sin(f*x+e)+35*B*cos(f*x+e)^2*sin(f*x+e)+56*A*cos(f*x+e)^2+8*B*cos(f*x+e)^2+35*A*s
in(f*x+e)+35*B*sin(f*x+e)+112*A+16*B)*(-c*(-1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(7/2)/(1+sin(f*
x+e))/cos(f*x+e)^5

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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{5}{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))^(5/2),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)^(5/2), x)

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Fricas [A]  time = 1.73989, size = 373, normalized size = 1.94 \begin{align*} -\frac{{\left (35 \,{\left (A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{6} - 35 \,{\left (A + B\right )} a^{3} c^{2} + 2 \,{\left (15 \, B a^{3} c^{2} \cos \left (f x + e\right )^{6} - 3 \,{\left (7 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{4} - 4 \,{\left (7 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{2} - 8 \,{\left (7 \, A + B\right )} a^{3} c^{2}\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}}{210 \, 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))^(5/2),x, algorithm="fricas")

[Out]

-1/210*(35*(A + B)*a^3*c^2*cos(f*x + e)^6 - 35*(A + B)*a^3*c^2 + 2*(15*B*a^3*c^2*cos(f*x + e)^6 - 3*(7*A + B)*
a^3*c^2*cos(f*x + e)^4 - 4*(7*A + B)*a^3*c^2*cos(f*x + e)^2 - 8*(7*A + B)*a^3*c^2)*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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^(5/2),x, algorithm="giac")

[Out]

sage2