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

Optimal. Leaf size=96 \[ \frac{c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{B c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt{c-c \sin (e+f x)}} \]

[Out]

((A - B)*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (B*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(7/2))/(4*a*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.318236, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {2971, 2738} \[ \frac{c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{B c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

((A - B)*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (B*c*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(7/2))/(4*a*f*Sqrt[c - c*Sin[e + f*x]])

Rule 2971

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f
, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 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 (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx &=\frac{B \int (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)} \, dx}{a}-(-A+B) \int (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{B c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 a f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.874536, size = 102, normalized size = 1.06 \[ \frac{a^2 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (16 (7 A+2 B) \sin (e+f x)-4 \cos (2 (e+f x)) (4 (A+2 B) \sin (e+f x)+12 A+9 B)+3 B \cos (4 (e+f x)))}{96 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(3*B*Cos[4*(e + f*x)] + 16*(7*A + 2*B)*S
in[e + f*x] - 4*Cos[2*(e + f*x)]*(12*A + 9*B + 4*(A + 2*B)*Sin[e + f*x])))/(96*f)

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Maple [A]  time = 0.339, size = 129, normalized size = 1.3 \begin{align*}{\frac{ \left ( 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}+8\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-12\,A\sin \left ( fx+e \right ) -9\,B\sin \left ( fx+e \right ) -16\,A-8\,B \right ) \sin \left ( fx+e \right ) }{12\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sin \left ( fx+e \right ) -2 \right ) \cos \left ( fx+e \right ) }\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{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))^(1/2),x)

[Out]

1/12/f*(3*B*cos(f*x+e)^2*sin(f*x+e)+4*A*cos(f*x+e)^2+8*B*cos(f*x+e)^2-12*A*sin(f*x+e)-9*B*sin(f*x+e)-16*A-8*B)
*(-c*(-1+sin(f*x+e)))^(1/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/(cos(f*x+e)^2-2*sin(f*x+e)-2)/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{-c \sin \left (f x + e\right ) + c}\,{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))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)*sqrt(-c*sin(f*x + e) + c), x)

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Fricas [A]  time = 1.95623, size = 293, normalized size = 3.05 \begin{align*} \frac{{\left (3 \, B a^{2} \cos \left (f x + e\right )^{4} - 12 \,{\left (A + B\right )} a^{2} \cos \left (f x + e\right )^{2} + 3 \,{\left (4 \, A + 3 \, B\right )} a^{2} - 4 \,{\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (2 \, A + 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}}{12 \, f \cos \left (f x + e\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))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*B*a^2*cos(f*x + e)^4 - 12*(A + B)*a^2*cos(f*x + e)^2 + 3*(4*A + 3*B)*a^2 - 4*((A + 2*B)*a^2*cos(f*x +
e)^2 - 2*(2*A + B)*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(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))**(1/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError