3.410 \(\int \frac{(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=86 \[ \frac{2^{m+\frac{1}{2}} \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac{5}{2},\frac{1}{2}-m;-\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 a^2 c^3 f} \]

[Out]

(2^(1/2 + m)*Hypergeometric2F1[-5/2, 1/2 - m, -3/2, (1 - Sin[e + f*x])/2]*Sec[e + f*x]^5*(1 + Sin[e + f*x])^(1
/2 - m)*(a + a*Sin[e + f*x])^(2 + m))/(5*a^2*c^3*f)

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Rubi [A]  time = 0.133367, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2689, 70, 69} \[ \frac{2^{m+\frac{1}{2}} \sec ^5(e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^{m+2} \, _2F_1\left (-\frac{5}{2},\frac{1}{2}-m;-\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{5 a^2 c^3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m/(c - c*Sin[e + f*x])^3,x]

[Out]

(2^(1/2 + m)*Hypergeometric2F1[-5/2, 1/2 - m, -3/2, (1 - Sin[e + f*x])/2]*Sec[e + f*x]^5*(1 + Sin[e + f*x])^(1
/2 - m)*(a + a*Sin[e + f*x])^(2 + m))/(5*a^2*c^3*f)

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^3} \, dx &=\frac{\int \sec ^6(e+f x) (a+a \sin (e+f x))^{3+m} \, dx}{a^3 c^3}\\ &=\frac{\left (\sec ^5(e+f x) (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{(a-a x)^{7/2}} \, dx,x,\sin (e+f x)\right )}{a c^3 f}\\ &=\frac{\left (2^{-\frac{1}{2}+m} \sec ^5(e+f x) (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{2+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{\frac{1}{2}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{1}{2}+m}}{(a-a x)^{7/2}} \, dx,x,\sin (e+f x)\right )}{a c^3 f}\\ &=\frac{2^{\frac{1}{2}+m} \, _2F_1\left (-\frac{5}{2},\frac{1}{2}-m;-\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) \sec ^5(e+f x) (1+\sin (e+f x))^{\frac{1}{2}-m} (a+a \sin (e+f x))^{2+m}}{5 a^2 c^3 f}\\ \end{align*}

Mathematica [C]  time = 23.3967, size = 7184, normalized size = 83.53 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m/(c - c*Sin[e + f*x])^3,x]

[Out]

Result too large to show

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Maple [F]  time = 0.368, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}}{ \left ( c-c\sin \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^3,x)

[Out]

int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

integral(-(a*sin(f*x + e) + a)^m/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x
)

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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))**m/(c-c*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c)^3, x)