3.157 \(\int \frac{(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{(A-7 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 7*B)*Cos[e + f*x]*(
a + a*Sin[e + f*x])^(5/2))/(48*c*f*(c - c*Sin[e + f*x])^(7/2))

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Rubi [A]  time = 0.275696, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {2972, 2742} \[ \frac{(A-7 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}} \]

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])^(9/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 7*B)*Cos[e + f*x]*(
a + a*Sin[e + f*x])^(5/2))/(48*c*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 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))^{9/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac{(A-7 B) \int \frac{(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{8 c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac{(A-7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 3.0597, size = 145, normalized size = 1.51 \[ \frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) ((4 A+17 B) \sin (e+f x)-3 (A-B) \cos (2 (e+f x))+5 A-3 B \sin (3 (e+f x))-5 B)}{12 c^4 f (\sin (e+f x)-1)^4 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

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])^(9/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(5*A - 5*B - 3*(A - B)*Cos[2*(e + f*x)]
+ (4*A + 17*B)*Sin[e + f*x] - 3*B*Sin[3*(e + f*x)]))/(12*c^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin
[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 0.283, size = 309, normalized size = 3.2 \begin{align*} -{\frac{ \left ( A \left ( \cos \left ( fx+e \right ) \right ) ^{4}-A \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) -B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+B \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) +4\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}+5\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}+B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -9\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -10\,A\cos \left ( fx+e \right ) -14\,A\sin \left ( fx+e \right ) -2\,B\cos \left ( fx+e \right ) +2\,B\sin \left ( fx+e \right ) +14\,A-2\,B \right ) \sin \left ( fx+e \right ) }{6\,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{9}{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))^(9/2),x)

[Out]

-1/6/f*(A*cos(f*x+e)^4-A*cos(f*x+e)^3*sin(f*x+e)-B*cos(f*x+e)^4+B*cos(f*x+e)^3*sin(f*x+e)+4*A*cos(f*x+e)^3+5*A
*cos(f*x+e)^2*sin(f*x+e)+2*B*cos(f*x+e)^3+B*cos(f*x+e)^2*sin(f*x+e)-9*A*cos(f*x+e)^2+4*A*sin(f*x+e)*cos(f*x+e)
+3*B*cos(f*x+e)^2-4*B*sin(f*x+e)*cos(f*x+e)-10*A*cos(f*x+e)-14*A*sin(f*x+e)-2*B*cos(f*x+e)+2*B*sin(f*x+e)+14*A
-2*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)*co
s(f*x+e)-2*cos(f*x+e)+4*sin(f*x+e)+4)/(-c*(-1+sin(f*x+e)))^(9/2)

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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))^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.50729, size = 400, normalized size = 4.17 \begin{align*} -\frac{{\left (3 \,{\left (A - B\right )} a^{2} \cos \left (f x + e\right )^{2} - 4 \,{\left (A - B\right )} a^{2} + 2 \,{\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} -{\left (A + 5 \, 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}}{6 \,{\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \,{\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} 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))^(9/2),x, algorithm="fricas")

[Out]

-1/6*(3*(A - B)*a^2*cos(f*x + e)^2 - 4*(A - B)*a^2 + 2*(3*B*a^2*cos(f*x + e)^2 - (A + 5*B)*a^2)*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^5*f*cos(f*x + e)^5 - 8*c^5*f*cos(f*x + e)^3 + 8*c^5*f*co
s(f*x + e) + 4*(c^5*f*cos(f*x + e)^3 - 2*c^5*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))**(9/2),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{9}{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))^(9/2),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)^(9/2), x)