3.146 \(\int \frac{(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x])^(7/2)) + ((A - 5*B)*Cos[e + f*x]*(
a + a*Sin[e + f*x])^(3/2))/(24*c*f*(c - c*Sin[e + f*x])^(5/2))

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Rubi [A]  time = 0.274596, 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} \[ \frac{(A-5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - c*Sin[e + f*x])^(7/2)) + ((A - 5*B)*Cos[e + f*x]*(
a + a*Sin[e + f*x])^(3/2))/(24*c*f*(c - c*Sin[e + f*x])^(5/2))

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))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac{(A-5 B) \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{6 c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}+\frac{(A-5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.03821, size = 125, normalized size = 1.3 \[ -\frac{a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (3 (A-B) \sin (e+f x)+A-3 B \cos (2 (e+f x))+4 B)}{6 c^3 f (\sin (e+f x)-1)^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(A + 4*B - 3*B*Cos[2*(e + f*x)] + 3*(A -
B)*Sin[e + f*x]))/(6*c^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]
])

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Maple [B]  time = 0.273, size = 223, normalized size = 2.3 \begin{align*}{\frac{ \left ( A \left ( \cos \left ( fx+e \right ) \right ) ^{3}+A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +B \left ( \cos \left ( fx+e \right ) \right ) ^{3}+B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -4\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3\,A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\,B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -7\,A\cos \left ( fx+e \right ) -10\,A\sin \left ( fx+e \right ) -B\cos \left ( fx+e \right ) +2\,B\sin \left ( fx+e \right ) +10\,A-2\,B \right ) \sin \left ( fx+e \right ) }{6\,f \left ( \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) +\cos \left ( fx+e \right ) -2 \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( -c \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))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x)

[Out]

1/6/f*(A*cos(f*x+e)^3+A*cos(f*x+e)^2*sin(f*x+e)+B*cos(f*x+e)^3+B*cos(f*x+e)^2*sin(f*x+e)-4*A*cos(f*x+e)^2+3*A*
sin(f*x+e)*cos(f*x+e)+2*B*cos(f*x+e)^2-3*B*sin(f*x+e)*cos(f*x+e)-7*A*cos(f*x+e)-10*A*sin(f*x+e)-B*cos(f*x+e)+2
*B*sin(f*x+e)+10*A-2*B)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)/(sin(f*x+e)*cos(f*x+e)+cos(f*x+e)^2-2*sin(f*x+e)+c
os(f*x+e)-2)/(-c*(-1+sin(f*x+e)))^(7/2)

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Maxima [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{3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(7/2), x)

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Fricas [A]  time = 2.02924, size = 309, normalized size = 3.22 \begin{align*} \frac{{\left (6 \, B a \cos \left (f x + e\right )^{2} - 3 \,{\left (A - B\right )} a \sin \left (f x + e\right ) -{\left (A + 7 \, B\right )} a\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{6 \,{\left (3 \, c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} f \cos \left (f x + e\right ) -{\left (c^{4} f \cos \left (f x + e\right )^{3} - 4 \, c^{4} 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))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/6*(6*B*a*cos(f*x + e)^2 - 3*(A - B)*a*sin(f*x + e) - (A + 7*B)*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x +
e) + c)/(3*c^4*f*cos(f*x + e)^3 - 4*c^4*f*cos(f*x + e) - (c^4*f*cos(f*x + e)^3 - 4*c^4*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))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),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{3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(7/2), x)