3.23 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=230 \[ -\frac {\left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{32 c^2 f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/8*(A+B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(c-c*sin(f*x+e))^(5/2)+1/16*(A*(5-2*m)-B*(3+2*m)-C*(11+2*m))
*cos(f*x+e)*(a+a*sin(f*x+e))^m/c/f/(c-c*sin(f*x+e))^(3/2)-1/32*(B*(-4*m^2-8*m+5)-A*(4*m^2-8*m+3)-C*(4*m^2+24*m
+19))*cos(f*x+e)*hypergeom([1, 1/2+m],[3/2+m],1/2+1/2*sin(f*x+e))*(a+a*sin(f*x+e))^m/c^2/f/(1+2*m)/(c-c*sin(f*
x+e))^(1/2)

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Rubi [A]  time = 0.70, antiderivative size = 230, normalized size of antiderivative = 1.00, 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, 2972, 2745, 2667, 68} \[ -\frac {\left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{32 c^2 f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}} \]

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])^(5/2),x]

[Out]

((A + B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(8*a*f*(c - c*Sin[e + f*x])^(5/2)) + ((A*(5 - 2*m) - B
*(3 + 2*m) - C*(11 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(16*c*f*(c - c*Sin[e + f*x])^(3/2)) - ((B*(5 -
8*m - 4*m^2) - A*(3 - 8*m + 4*m^2) - C*(19 + 24*m + 4*m^2))*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)/(32*c^2*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

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]

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 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 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 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]))

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))^{5/2}} \, dx &=\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}-\frac {\int \frac {(a+a \sin (e+f x))^m \left (-\frac {1}{2} a^2 (A (9-2 m)-(B+C) (7+2 m))-\frac {1}{2} a^2 ((A+B) (1-2 m)-C (15+2 m)) \sin (e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 a^2 c}\\ &=\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac {(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx}{32 c^2}\\ &=\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac {(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x)\right ) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac {1}{2}+m} \, dx}{32 c^2 \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}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac {(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a \left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-\frac {1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{32 c^2 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}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac {(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \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}{32 c^2 f (1+2 m) \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 7.32, size = 8321, normalized size = 36.18 \[ \text {Result too large to show} \]

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])^(5/2),x]

[Out]

Result too large to show

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

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))^(5/2),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/(3*c^3*c
os(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),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)^(5/2), x)

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maple [F]  time = 4.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx \]

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))^(5/2),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))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),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)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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))^(5/2),x)

[Out]

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))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(5/2),x)

[Out]

Timed out

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