3.171 \(\int \frac{(a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{480 c^2 f (c-c \sin (e+f x))^{9/2}}+\frac{(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{60 c f (c-c \sin (e+f x))^{11/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{12 f (c-c \sin (e+f x))^{13/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(12*f*(c - c*Sin[e + f*x])^(13/2)) + ((A - 5*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(7/2))/(60*c*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 5*B)*Cos[e + f*x]*(a + a*Sin[e + f*x
])^(7/2))/(480*c^2*f*(c - c*Sin[e + f*x])^(9/2))

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Rubi [A]  time = 0.378984, 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} \[ \frac{(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{480 c^2 f (c-c \sin (e+f x))^{9/2}}+\frac{(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{60 c f (c-c \sin (e+f x))^{11/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{12 f (c-c \sin (e+f x))^{13/2}} \]

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])^(13/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(12*f*(c - c*Sin[e + f*x])^(13/2)) + ((A - 5*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(7/2))/(60*c*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 5*B)*Cos[e + f*x]*(a + a*Sin[e + f*x
])^(7/2))/(480*c^2*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 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))^{7/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac{(A-5 B) \int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{11/2}} \, dx}{6 c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac{(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{60 c f (c-c \sin (e+f x))^{11/2}}+\frac{(A-5 B) \int \frac{(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{60 c^2}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{12 f (c-c \sin (e+f x))^{13/2}}+\frac{(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{60 c f (c-c \sin (e+f x))^{11/2}}+\frac{(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{480 c^2 f (c-c \sin (e+f x))^{9/2}}\\ \end{align*}

Mathematica [B]  time = 6.94883, size = 442, normalized size = 3.03 \[ \frac{(-A-7 B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7}{3 f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{3 (A+3 B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5}{2 f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}-\frac{4 (3 A+5 B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}{5 f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{4 (A+B) (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{3 f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}+\frac{B (a (\sin (e+f x)+1))^{7/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9}{2 f (c-c \sin (e+f x))^{13/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

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])^(13/2),x]

[Out]

(4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(7/2))/(3*f*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^7*(c - c*Sin[e + f*x])^(13/2)) - (4*(3*A + 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a*(1 + Si
n[e + f*x]))^(7/2))/(5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(13/2)) + (3*(A + 3*B)*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(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])^(13/2)) + ((-A - 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(a*(1 + Sin[e + f*x]))^
(7/2))/(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(c - c*Sin[e + f*x])^(13/2)) + (B*(Cos[(e + f*x)/2] - Sin[
(e + f*x)/2])^9*(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]
)^(13/2))

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Maple [B]  time = 0.331, size = 393, normalized size = 2.7 \begin{align*}{\frac{ \left ( 3\,A \left ( \cos \left ( fx+e \right ) \right ) ^{6}-3\,A \left ( \cos \left ( fx+e \right ) \right ) ^{5}\sin \left ( fx+e \right ) +18\,A \left ( \cos \left ( fx+e \right ) \right ) ^{5}+21\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}\sin \left ( fx+e \right ) -72\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+51\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) +15\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}-15\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ) -106\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}-157\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -10\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}+5\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +235\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-78\,A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -35\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+30\,B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +118\,A\cos \left ( fx+e \right ) +196\,A\sin \left ( fx+e \right ) +10\,B\cos \left ( fx+e \right ) -20\,B\sin \left ( fx+e \right ) -196\,A+20\,B \right ) \sin \left ( fx+e \right ) }{30\,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{13}{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))^(13/2),x)

[Out]

1/30/f*(3*A*cos(f*x+e)^6-3*A*cos(f*x+e)^5*sin(f*x+e)+18*A*cos(f*x+e)^5+21*A*cos(f*x+e)^4*sin(f*x+e)-72*A*cos(f
*x+e)^4+51*A*cos(f*x+e)^3*sin(f*x+e)+15*B*cos(f*x+e)^4-15*B*cos(f*x+e)^3*sin(f*x+e)-106*A*cos(f*x+e)^3-157*A*c
os(f*x+e)^2*sin(f*x+e)-10*B*cos(f*x+e)^3+5*B*cos(f*x+e)^2*sin(f*x+e)+235*A*cos(f*x+e)^2-78*A*sin(f*x+e)*cos(f*
x+e)-35*B*cos(f*x+e)^2+30*B*sin(f*x+e)*cos(f*x+e)+118*A*cos(f*x+e)+196*A*sin(f*x+e)+10*B*cos(f*x+e)-20*B*sin(f
*x+e)-196*A+20*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))
)^(13/2)

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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))^(13/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.28966, size = 535, normalized size = 3.66 \begin{align*} -\frac{{\left (15 \, B a^{3} \cos \left (f x + e\right )^{4} - 15 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 6 \,{\left (3 \, A + 5 \, B\right )} a^{3} - 2 \,{\left (5 \,{\left (A + B\right )} a^{3} \cos \left (f x + e\right )^{2} -{\left (11 \, A + 5 \, 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}}{30 \,{\left (c^{7} f \cos \left (f x + e\right )^{7} - 18 \, c^{7} f \cos \left (f x + e\right )^{5} + 48 \, c^{7} f \cos \left (f x + e\right )^{3} - 32 \, c^{7} f \cos \left (f x + e\right ) + 2 \,{\left (3 \, c^{7} f \cos \left (f x + e\right )^{5} - 16 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} 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))^(13/2),x, algorithm="fricas")

[Out]

-1/30*(15*B*a^3*cos(f*x + e)^4 - 15*(A + 3*B)*a^3*cos(f*x + e)^2 + 6*(3*A + 5*B)*a^3 - 2*(5*(A + B)*a^3*cos(f*
x + e)^2 - (11*A + 5*B)*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^7*f*cos(f*x +
e)^7 - 18*c^7*f*cos(f*x + e)^5 + 48*c^7*f*cos(f*x + e)^3 - 32*c^7*f*cos(f*x + e) + 2*(3*c^7*f*cos(f*x + e)^5
- 16*c^7*f*cos(f*x + e)^3 + 16*c^7*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))**(13/2),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{13}{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))^(13/2),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)^(13/2), x)