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

Optimal. Leaf size=216 \[ \frac{(A (1-2 m)-B (2 m+3)-C (2 m+7)) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(2 A m+A+2 B m+B+C (2 m+9)) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 c f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{4 a f (c-c \sin (e+f x))^{3/2}} \]

[Out]

((A + B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(4*a*f*(c - c*Sin[e + f*x])^(3/2)) + ((A + B + 2*A*m +
2*B*m + C*(9 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(4*c*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + ((A*(1
- 2*m) - B*(3 + 2*m) - C*(7 + 2*m))*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]
*(a + a*Sin[e + f*x])^m)/(4*c*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.677942, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.104, Rules used = {3035, 2973, 2745, 2667, 68} \[ \frac{(A (1-2 m)-B (2 m+3)-C (2 m+7)) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(2 A m+A+2 B m+B+C (2 m+9)) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 c f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{4 a f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(4*a*f*(c - c*Sin[e + f*x])^(3/2)) + ((A + B + 2*A*m +
2*B*m + C*(9 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(4*c*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + ((A*(1
- 2*m) - B*(3 + 2*m) - C*(7 + 2*m))*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]
*(a + a*Sin[e + f*x])^m)/(4*c*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

Rule 3035

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(2*b*c*f*(2*m + 1)), x] - Dist[1/(2*b*c*d*(2*m + 1)), Int[
(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n -
1) - C*(c^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (LtQ[m, -2^(-1)] || (EqQ[m +
n + 2, 0] && NeQ[2*m + 1, 0]))

Rule 2973

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((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)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(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)] && NeQ[m + n + 1, 0]

Rule 2745

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^FracPart[m])/Cos[e + f*x]^(2
*FracPart[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, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

Rule 2667

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\int \frac{(a+a \sin (e+f x))^m \left (-\frac{1}{2} a^2 (A (3-2 m)-(B+C) (5+2 m))+\frac{1}{2} a^2 (A+B+2 A m+2 B m+C (9+2 m)) \sin (e+f x)\right )}{\sqrt{c-c \sin (e+f x)}} \, dx}{4 a^2 c}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B+2 A m+2 B m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{(A (1-2 m)-B (3+2 m)-C (7+2 m)) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c-c \sin (e+f x)}} \, dx}{4 c}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B+2 A m+2 B m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{((A (1-2 m)-B (3+2 m)-C (7+2 m)) \cos (e+f x)) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac{1}{2}+m} \, dx}{4 c \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B+2 A m+2 B m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{(a (A (1-2 m)-B (3+2 m)-C (7+2 m)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+x)^{-\frac{1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{4 c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B+2 A m+2 B m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt{c-c \sin (e+f x)}}+\frac{(A (1-2 m)-B (3+2 m)-C (7+2 m)) \cos (e+f x) \, _2F_1\left (1,\frac{1}{2}+m;\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 25.012, size = 14053, normalized size = 65.06 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

Maple [F]  time = 0.734, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*(A+B*sin(f*x+e)+C*sin(f*x+e)**2)/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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