3.170 \(\int \frac{(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{(A-9 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 9*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(7/2))/(80*c*f*(c - c*Sin[e + f*x])^(9/2))

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Rubi [A]  time = 0.271934, 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} \[ \frac{(A-9 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

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])^(11/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 9*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(7/2))/(80*c*f*(c - c*Sin[e + f*x])^(9/2))

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))^{7/2} (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))^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{(A-9 B) \int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{10 c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{(A-9 B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}}\\ \end{align*}

Mathematica [B]  time = 6.91147, size = 434, normalized size = 4.52 \[ \frac{(-A-7 B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7}{2 f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{2 (A+3 B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5}{f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{(-3 A-5 B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}{f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{8 (A+B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{5 f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{B (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9}{f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

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])^(11/2),x]

[Out]

(8*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2))/(5*f*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2)) + ((-3*A - 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a*(1 + Sin
[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2)) + (2*(A + 3*B)*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a*(1 + Sin[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(
c - c*Sin[e + f*x])^(11/2)) + ((-A - 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(a*(1 + Sin[e + f*x]))^(7/2)
)/(2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2)) + (B*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(7/2))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(11/2
))

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Maple [B]  time = 0.298, size = 389, normalized size = 4.1 \begin{align*}{\frac{ \left ( A \left ( \cos \left ( fx+e \right ) \right ) ^{5}+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}-6\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+5\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) +4\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}-5\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) -17\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}-22\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -7\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}-2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +32\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-10\,A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -8\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+10\,B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +26\,A\cos \left ( fx+e \right ) +36\,A\sin \left ( fx+e \right ) +6\,B\cos \left ( fx+e \right ) -4\,B\sin \left ( fx+e \right ) -36\,A+4\,B \right ) \sin \left ( fx+e \right ) }{10\,f \left ( \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}+ \left ( \cos \left ( fx+e \right ) \right ) ^{4}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-4\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\sin \left ( fx+e \right ) -4\,\cos \left ( fx+e \right ) +8 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{7}{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))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x)

[Out]

1/10/f*(A*cos(f*x+e)^5+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)^4-6*A*cos(f*x+e)^4+5*A
*cos(f*x+e)^3*sin(f*x+e)+4*B*cos(f*x+e)^4-5*B*cos(f*x+e)^3*sin(f*x+e)-17*A*cos(f*x+e)^3-22*A*cos(f*x+e)^2*sin(
f*x+e)-7*B*cos(f*x+e)^3-2*B*cos(f*x+e)^2*sin(f*x+e)+32*A*cos(f*x+e)^2-10*A*sin(f*x+e)*cos(f*x+e)-8*B*cos(f*x+e
)^2+10*B*sin(f*x+e)*cos(f*x+e)+26*A*cos(f*x+e)+36*A*sin(f*x+e)+6*B*cos(f*x+e)-4*B*sin(f*x+e)-36*A+4*B)*sin(f*x
+e)*(a*(1+sin(f*x+e)))^(7/2)/(sin(f*x+e)*cos(f*x+e)^3+cos(f*x+e)^4-4*cos(f*x+e)^2*sin(f*x+e)+3*cos(f*x+e)^3-4*
sin(f*x+e)*cos(f*x+e)-8*cos(f*x+e)^2+8*sin(f*x+e)-4*cos(f*x+e)+8)/(-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))^(7/2)*(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 [B]  time = 2.41408, size = 489, normalized size = 5.09 \begin{align*} \frac{{\left (10 \, B a^{3} \cos \left (f x + e\right )^{4} - 5 \,{\left (A + 7 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 2 \,{\left (3 \, A + 13 \, B\right )} a^{3} - 5 \,{\left ({\left (A - B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \,{\left (A - B\right )} a^{3}\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}}{10 \,{\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))^(7/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

1/10*(10*B*a^3*cos(f*x + e)^4 - 5*(A + 7*B)*a^3*cos(f*x + e)^2 + 2*(3*A + 13*B)*a^3 - 5*((A - B)*a^3*cos(f*x +
e)^2 - 2*(A - B)*a^3)*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))**(7/2)*(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{7}{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))^(7/2)*(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)^(7/2)/(-c*sin(f*x + e) + c)^(11/2), x)