3.10 \(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (A+C \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=392 \[ -\frac{a 2^{m+\frac{1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)+c^2 (-(2 C m+C))\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac{(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{d f (m+1) (c-d) (c+d)^2}+\frac{\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}+\frac{\sqrt{2} C \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \left (\frac{c+d \sin (e+f x)}{c-d}\right )^m (c+d \sin (e+f x))^{-m} F_1\left (m+\frac{3}{2};\frac{1}{2},m+1;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (c-d) \sqrt{1-\sin (e+f x)}} \]

[Out]

((c^2*C + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m))/(d*(c^2 - d^2)*f*(1 + m))
- (2^(1/2 + m)*a*(c*(A + C)*d*(1 + m) + d^2*(C - A*m + C*m) - c^2*(C + 2*C*m))*Cos[e + f*x]*Hypergeometric2F1[
1/2, 1/2 - m, 3/2, ((c - d)*(1 - Sin[e + f*x]))/(2*(c + d*Sin[e + f*x]))]*(a + a*Sin[e + f*x])^(-1 + m)*(((c +
d)*(1 + Sin[e + f*x]))/(c + d*Sin[e + f*x]))^(1/2 - m))/((c - d)*d*(c + d)^2*f*(1 + m)*(c + d*Sin[e + f*x])^m
) + (Sqrt[2]*C*AppellF1[3/2 + m, 1/2, 1 + m, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]
*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*((c + d*Sin[e + f*x])/(c - d))^m)/(a*(c - d)*d*f*(3 + 2*m)*Sqrt[1 -
Sin[e + f*x]]*(c + d*Sin[e + f*x])^m)

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Rubi [A]  time = 0.99164, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3044, 2987, 2788, 132, 140, 139, 138} \[ -\frac{a 2^{m+\frac{1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)+c^2 (-(2 C m+C))\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac{(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{d f (m+1) (c-d) (c+d)^2}+\frac{\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}+\frac{\sqrt{2} C \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \left (\frac{c+d \sin (e+f x)}{c-d}\right )^m (c+d \sin (e+f x))^{-m} F_1\left (m+\frac{3}{2};\frac{1}{2},m+1;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (c-d) \sqrt{1-\sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-2 - m)*(A + C*Sin[e + f*x]^2),x]

[Out]

((c^2*C + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m))/(d*(c^2 - d^2)*f*(1 + m))
- (2^(1/2 + m)*a*(c*(A + C)*d*(1 + m) + d^2*(C - A*m + C*m) - c^2*(C + 2*C*m))*Cos[e + f*x]*Hypergeometric2F1[
1/2, 1/2 - m, 3/2, ((c - d)*(1 - Sin[e + f*x]))/(2*(c + d*Sin[e + f*x]))]*(a + a*Sin[e + f*x])^(-1 + m)*(((c +
d)*(1 + Sin[e + f*x]))/(c + d*Sin[e + f*x]))^(1/2 - m))/((c - d)*d*(c + d)^2*f*(1 + m)*(c + d*Sin[e + f*x])^m
) + (Sqrt[2]*C*AppellF1[3/2 + m, 1/2, 1 + m, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]
*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*((c + d*Sin[e + f*x])/(c - d))^m)/(a*(c - d)*d*f*(3 + 2*m)*Sqrt[1 -
Sin[e + f*x]]*(c + d*Sin[e + f*x])^m)

Rule 3044

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

Rule 2987

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x
], x] + Dist[B/b, 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] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dis
t[(a^2*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[a - b*x], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] &
& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !IntegerQ[m]

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx &=\frac{\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac{\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \left (-a (A d (c+c m-d m)+c C (d-c m+d m))-a C \left (c^2-d^2\right ) (1+m) \sin (e+f x)\right ) \, dx}{a d \left (c^2-d^2\right ) (1+m)}\\ &=\frac{\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}+\frac{C \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-1-m} \, dx}{a d}+\frac{\left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx}{d \left (c^2-d^2\right ) (1+m)}\\ &=\frac{\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}+\frac{(a C \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^{-1-m}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^{-1-m}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d \left (c^2-d^2\right ) f (1+m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac{2^{\frac{1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac{\left (a C \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^{-1-m}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac{2^{\frac{1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac{\left (a^2 C \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} (c+d \sin (e+f x))^{-m} \left (\frac{a (c+d \sin (e+f x))}{a c-a d}\right )^m\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^{-1-m}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d (a c-a d) f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=\frac{\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac{2^{\frac{1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac{(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac{1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac{\sqrt{2} C F_1\left (\frac{3}{2}+m;\frac{1}{2},1+m;\frac{5}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt{1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac{c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) d f (3+2 m) (a-a \sin (e+f x))}\\ \end{align*}

Mathematica [B]  time = 57.0758, size = 7563, normalized size = 19.29 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-2 - m)*(A + C*Sin[e + f*x]^2),x]

[Out]

Result too large to show

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Maple [F]  time = 1.423, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{-2-m} \left ( A+C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x)

[Out]

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x)

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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))^m*(c+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral(-(C*cos(f*x + e)^2 - A - C)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^(-m - 2), 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+d*sin(f*x+e))**(-2-m)*(A+C*sin(f*x+e)**2),x)

[Out]

Timed out

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Giac [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+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

Timed out