\(\int \frac {(a+a \sin (e+f x))^m (A+C \sin ^2(e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\) [5]

Optimal result

Integrand size = 40, antiderivative size = 202 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac {(A+2 A m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A (1-2 m)-C (7+2 m)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/4*(A+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(c-c*sin(f*x+e))^(3/2)+1/4*(A+2*A*m+C*(9+2*m))*cos(f*x+e)*(a+a
*sin(f*x+e))^m/c/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)+1/4*(A*(1-2*m)-C*(7+2*m))*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/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3115, 3052, 2824, 2746, 70} \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {(A (1-2 m)-C (2 m+7)) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(2 A m+A+C (2 m+9)) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(A+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{4 a f (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

((A + 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 + 2*A*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) - 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 70

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

Rule 2746

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 2824

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 3052

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*Si
n[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 3115

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[(a*A + 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*Si
n[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - C*(c^2*m - d^2*(n + 1)) + d*(A*c*(m + n + 2) - c*C*(3
*m - n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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*} \text {integral}& = \frac {(A+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)-C (5+2 m))+\frac {1}{2} a^2 (A+2 A m+C (9+2 m)) \sin (e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx}{4 a^2 c} \\ & = \frac {(A+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+2 A 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)-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+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+2 A 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)-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+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+2 A 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)-C (7+2 m)) \cos (e+f x)) \text {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+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+2 A 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)-C (7+2 m)) \cos (e+f x) \operatorname {Hypergeometric2F1}\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 [A] (verified)

Time = 52.52 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.54 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\cos (e+f x) \left (-4 C+4 C \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )-(A+C) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )\right ) (a (1+\sin (e+f x)))^m}{2 c f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

-1/2*(Cos[e + f*x]*(-4*C + 4*C*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2] - (A + C)*Hypergeo
metric2F1[2, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2])*(a*(1 + Sin[e + f*x]))^m)/(c*f*(1 + 2*m)*Sqrt[c - c*Sin[
e + f*x]])

Maple [F]

\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

[In]

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

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+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 - 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]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + C*sin(e + f*x)**2)/(-c*(sin(e + f*x) - 1))**(3/2), x)

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{1,[0,1,1,1,0,0,0,0,0]%%%}+%%%{1,[0,0,1,1,1,0,0,0,0]%%%}
/ %%%{16,[0

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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