3.148 \(\int \frac{(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=154 \[ -\frac{a^2 (3 A-7 B) \cos (e+f x)}{120 c^2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}+\frac{a (3 A-7 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + (a*(3*A - 7*B)*Cos[e +
f*x]*Sqrt[a + a*Sin[e + f*x]])/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(3*A - 7*B)*Cos[e + f*x])/(120*c^2*f
*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.372795, antiderivative size = 154, 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, 2739, 2738} \[ -\frac{a^2 (3 A-7 B) \cos (e+f x)}{120 c^2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}+\frac{a (3 A-7 B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

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])^(11/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + (a*(3*A - 7*B)*Cos[e +
f*x]*Sqrt[a + a*Sin[e + f*x]])/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(3*A - 7*B)*Cos[e + f*x])/(120*c^2*f
*Sqrt[a + a*Sin[e + f*x]]*(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 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -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))^{11/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{(3 A-7 B) \int \frac{(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{10 c}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{a (3 A-7 B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{40 c f (c-c \sin (e+f x))^{9/2}}-\frac{(a (3 A-7 B)) \int \frac{\sqrt{a+a \sin (e+f x)}}{(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))^{3/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac{a (3 A-7 B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{40 c f (c-c \sin (e+f x))^{9/2}}-\frac{a^2 (3 A-7 B) \cos (e+f x)}{120 c^2 f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.96955, size = 126, normalized size = 0.82 \[ -\frac{a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (5 (3 A+B) \sin (e+f x)+9 (A+B)-10 B \cos (2 (e+f x)))}{60 c^5 f (\sin (e+f x)-1)^5 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(9*(A + B) - 10*B*Cos[2*(e + f*x)] + 5*(3
*A + B)*Sin[e + f*x]))/(60*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]])

________________________________________________________________________________________

Maple [B]  time = 0.299, size = 339, normalized size = 2.2 \begin{align*} -{\frac{ \left ( 9\,A \left ( \cos \left ( fx+e \right ) \right ) ^{5}+9\,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}-54\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}+45\,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 ) -108\,A \left ( \cos \left ( fx+e \right ) \right ) ^{3}-153\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +12\,B \left ( \cos \left ( fx+e \right ) \right ) ^{3}+17\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +288\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-135\,A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -52\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+35\,B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +159\,A\cos \left ( fx+e \right ) +294\,A\sin \left ( fx+e \right ) -11\,B\cos \left ( fx+e \right ) -46\,B\sin \left ( fx+e \right ) -294\,A+46\,B \right ) \sin \left ( fx+e \right ) }{60\,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{11}{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))^(11/2),x)

[Out]

-1/60/f*(9*A*cos(f*x+e)^5+9*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)^4-54*A*cos(f*x+e)
^4+45*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)-108*A*cos(f*x+e)^3-153*A*cos(f*x+
e)^2*sin(f*x+e)+12*B*cos(f*x+e)^3+17*B*cos(f*x+e)^2*sin(f*x+e)+288*A*cos(f*x+e)^2-135*A*sin(f*x+e)*cos(f*x+e)-
52*B*cos(f*x+e)^2+35*B*sin(f*x+e)*cos(f*x+e)+159*A*cos(f*x+e)+294*A*sin(f*x+e)-11*B*cos(f*x+e)-46*B*sin(f*x+e)
-294*A+46*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)+cos(f*x+e)-2
)/(-c*(-1+sin(f*x+e)))^(11/2)

________________________________________________________________________________________

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{11}{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))^(11/2),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)^(11/2), x)

________________________________________________________________________________________

Fricas [A]  time = 2.09092, size = 393, normalized size = 2.55 \begin{align*} -\frac{{\left (20 \, B a \cos \left (f x + e\right )^{2} - 5 \,{\left (3 \, A + B\right )} a \sin \left (f x + e\right ) -{\left (9 \, A + 19 \, B\right )} a\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{60 \,{\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))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/60*(20*B*a*cos(f*x + e)^2 - 5*(3*A + B)*a*sin(f*x + e) - (9*A + 19*B)*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*s
in(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))

________________________________________________________________________________________

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))**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{11}{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))^(11/2),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)^(11/2), x)