3.158 \(\int \frac{(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{(A-4 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}}+\frac{(A-4 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 4*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(5/2))/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 4*B)*Cos[e + f*x]*(a + a*Sin[e + f*x]
)^(5/2))/(240*c^2*f*(c - c*Sin[e + f*x])^(7/2))

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Rubi [A]  time = 0.376271, antiderivative size = 146, 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 = {2972, 2743, 2742} \[ \frac{(A-4 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}}+\frac{(A-4 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 4*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(5/2))/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 4*B)*Cos[e + f*x]*(a + a*Sin[e + f*x]
)^(5/2))/(240*c^2*f*(c - c*Sin[e + f*x])^(7/2))

Rule 2972

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

Rule 2743

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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)/(a*f*(2*m + 1)), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), 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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rule 2742

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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)/(a*f*(2*m + 1)), x] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{(A-4 B) \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{5 c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A-4 B) \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{40 c^2}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 4.20428, size = 146, normalized size = 1. \[ \frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (-5 (8 A+13 B) \sin (e+f x)+10 (2 A+B) \cos (2 (e+f x))-36 A+15 B \sin (3 (e+f x))-6 B)}{120 c^5 f (\sin (e+f x)-1)^5 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(-36*A - 6*B + 10*(2*A + B)*Cos[2*(e + f
*x)] - 5*(8*A + 13*B)*Sin[e + f*x] + 15*B*Sin[3*(e + f*x)]))/(120*c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*
(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 0.305, size = 368, normalized size = 2.5 \begin{align*}{\frac{ \left ( 4\,A \left ( \cos \left ( fx+e \right ) \right ) ^{5}+4\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}\sin \left ( fx+e \right ) -B \left ( \cos \left ( fx+e \right ) \right ) ^{5}-B\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-24\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+20\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) +6\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}-5\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) -48\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}-68\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}+2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +118\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-50\,A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -22\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+20\,B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +74\,A\cos \left ( fx+e \right ) +124\,A\sin \left ( fx+e \right ) +4\,B\cos \left ( fx+e \right ) -16\,B\sin \left ( fx+e \right ) -124\,A+16\,B \right ) \sin \left ( fx+e \right ) }{30\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{3}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -2\,\cos \left ( fx+e \right ) +4\,\sin \left ( fx+e \right ) +4 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x)

[Out]

1/30/f*(4*A*cos(f*x+e)^5+4*A*cos(f*x+e)^4*sin(f*x+e)-B*cos(f*x+e)^5-B*sin(f*x+e)*cos(f*x+e)^4-24*A*cos(f*x+e)^
4+20*A*cos(f*x+e)^3*sin(f*x+e)+6*B*cos(f*x+e)^4-5*B*cos(f*x+e)^3*sin(f*x+e)-48*A*cos(f*x+e)^3-68*A*cos(f*x+e)^
2*sin(f*x+e)-3*B*cos(f*x+e)^3+2*B*cos(f*x+e)^2*sin(f*x+e)+118*A*cos(f*x+e)^2-50*A*sin(f*x+e)*cos(f*x+e)-22*B*c
os(f*x+e)^2+20*B*sin(f*x+e)*cos(f*x+e)+74*A*cos(f*x+e)+124*A*sin(f*x+e)+4*B*cos(f*x+e)-16*B*sin(f*x+e)-124*A+1
6*B)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/(cos(f*x+e)^3-cos(f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2-2*sin(f*x+e)*cos
(f*x+e)-2*cos(f*x+e)+4*sin(f*x+e)+4)/(-c*(-1+sin(f*x+e)))^(11/2)

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Maxima [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))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.5766, size = 451, normalized size = 3.09 \begin{align*} -\frac{{\left (5 \,{\left (2 \, A + B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (7 \, A + 2 \, B\right )} a^{2} + 5 \,{\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + 2 \, B\right )} a^{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}}{30 \,{\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) -{\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/30*(5*(2*A + B)*a^2*cos(f*x + e)^2 - 2*(7*A + 2*B)*a^2 + 5*(3*B*a^2*cos(f*x + e)^2 - 2*(A + 2*B)*a^2)*sin(f
*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos(f*x + e)^3
+ 16*c^6*f*cos(f*x + e) - (c^6*f*cos(f*x + e)^5 - 12*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f*x + e))*sin(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))**(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(11/2), x)