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

Optimal. Leaf size=180 \[ \frac {2 c (C-6 C m+A (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {4 c C (1+2 m) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)} \]

[Out]

2*C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3/2)/c/f/(5+2*m)+2*c*(C-6*C*m+A*(5+2*m))*cos(f*x+e)*(a+a*s
in(f*x+e))^m/f/(4*m^2+12*m+5)/(c-c*sin(f*x+e))^(1/2)+4*c*C*(1+2*m)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(4*m^
2+16*m+15)/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.38, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3119, 3050, 2817} \begin {gather*} \frac {2 c (A (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+5) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}+\frac {4 c C (2 m+1) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) (2 m+5) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(C - 6*C*m + A*(5 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*(5 + 2*m)*Sqrt[c - c*Sin[e +
f*x]]) + (4*c*C*(1 + 2*m)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(3 + 2*m)*(5 + 2*m)*Sqrt[c - c*Sin[e
+ f*x]]) + (2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(c*f*(5 + 2*m))

Rule 2817

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

Rule 3050

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f
, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3119

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(
n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Si
mp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) - b*c*C*(2*m + 1)*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)] && NeQ[m + n + 2, 0]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+C \sin ^2(e+f x)\right ) \, dx &=\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}-\frac {2 \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (-\frac {1}{2} a c (C (3-2 m)+A (5+2 m))-a c C (1+2 m) \sin (e+f x)\right ) \, dx}{a c (5+2 m)}\\ &=\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}+\frac {(2 C (1+2 m)) \int (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)} \, dx}{a (5+2 m)}+\frac {(C-6 C m+A (5+2 m)) \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx}{5+2 m}\\ &=\frac {2 c (C-6 C m+A (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {4 c C (1+2 m) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 160, normalized size = 0.89 \begin {gather*} -\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)} \left (-30 A-19 C-32 A m-8 C m-8 A m^2-4 C m^2+C \left (3+8 m+4 m^2\right ) \cos (2 (e+f x))+8 C (1+2 m) \sin (e+f x)\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c*Sin[e + f*x]]*(-30*A - 19*C - 32*
A*m - 8*C*m - 8*A*m^2 - 4*C*m^2 + C*(3 + 8*m + 4*m^2)*Cos[2*(e + f*x)] + 8*C*(1 + 2*m)*Sin[e + f*x]))/(f*(1 +
2*m)*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])))

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Maple [F]
time = 1.17, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c -c \sin \left (f x +e \right )}\, \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx\]

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))^(1/2)*(A+C*sin(f*x+e)^2),x)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (179) = 358\).
time = 0.56, size = 469, normalized size = 2.61 \begin {gather*} \frac {2 \, {\left (\frac {4 \, {\left (\frac {4 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {4 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 2 \, a^{m} \sqrt {c} - \frac {2 \, a^{m} \sqrt {c} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} C e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + \frac {2 \, {\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac {{\left (a^{m} \sqrt {c} + \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} A e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (2 \, m + 1\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \end {gather*}

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))^(1/2)*(A+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

2*(4*(4*a^m*sqrt(c)*m*sin(f*x + e)/(cos(f*x + e) + 1) - (4*m^2 + 4*m + 5)*a^m*sqrt(c)*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 - (4*m^2 + 4*m + 5)*a^m*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 4*a^m*sqrt(c)*m*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 - 2*a^m*sqrt(c) - 2*a^m*sqrt(c)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*C*e^(2*m*log(sin(
f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^3 + 36*m^2 + 46*m + 2
*(8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^
4/(cos(f*x + e) + 1)^4 + 15)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)) - (a^m*sqrt(c) + a^m*sqrt(c)*sin(f
*x + e)/(cos(f*x + e) + 1))*A*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 1))/((2*m + 1)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)))/f

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Fricas [A]
time = 0.42, size = 271, normalized size = 1.51 \begin {gather*} -\frac {2 \, {\left ({\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 4 \, {\left (A + C\right )} m^{2} + {\left (4 \, C m^{2} - C\right )} \cos \left (f x + e\right )^{2} - 16 \, A m - {\left (4 \, {\left (A + C\right )} m^{2} + 8 \, {\left (2 \, A + C\right )} m + 15 \, A + 11 \, C\right )} \cos \left (f x + e\right ) - {\left (4 \, {\left (A + C\right )} m^{2} - {\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{2} + 16 \, A m - 4 \, {\left (2 \, C m + C\right )} \cos \left (f x + e\right ) + 15 \, A + 7 \, C\right )} \sin \left (f x + e\right ) - 15 \, A - 7 \, C\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \end {gather*}

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))^(1/2)*(A+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

-2*((4*C*m^2 + 8*C*m + 3*C)*cos(f*x + e)^3 - 4*(A + C)*m^2 + (4*C*m^2 - C)*cos(f*x + e)^2 - 16*A*m - (4*(A + C
)*m^2 + 8*(2*A + C)*m + 15*A + 11*C)*cos(f*x + e) - (4*(A + C)*m^2 - (4*C*m^2 + 8*C*m + 3*C)*cos(f*x + e)^2 +
16*A*m - 4*(2*C*m + C)*cos(f*x + e) + 15*A + 7*C)*sin(f*x + e) - 15*A - 7*C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(
f*x + e) + a)^m/(8*f*m^3 + 36*f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f*m^3 +
36*f*m^2 + 46*f*m + 15*f)*sin(f*x + e) + 15*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \end {gather*}

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))**(1/2)*(A+C*sin(f*x+e)**2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(1/2)*(A+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 17.16, size = 185, normalized size = 1.03 \begin {gather*} -\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (60\,A\,\cos \left (e+f\,x\right )+35\,C\,\cos \left (e+f\,x\right )-3\,C\,\cos \left (3\,e+3\,f\,x\right )-8\,C\,\sin \left (2\,e+2\,f\,x\right )-4\,C\,m^2\,\cos \left (3\,e+3\,f\,x\right )+64\,A\,m\,\cos \left (e+f\,x\right )+8\,C\,m\,\cos \left (e+f\,x\right )+16\,A\,m^2\,\cos \left (e+f\,x\right )-8\,C\,m\,\cos \left (3\,e+3\,f\,x\right )+4\,C\,m^2\,\cos \left (e+f\,x\right )-16\,C\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{2\,f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (8\,m^3+36\,m^2+46\,m+15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2)*(60*A*cos(e + f*x) + 35*C*cos(e + f*x) - 3*C*cos(3*e
+ 3*f*x) - 8*C*sin(2*e + 2*f*x) - 4*C*m^2*cos(3*e + 3*f*x) + 64*A*m*cos(e + f*x) + 8*C*m*cos(e + f*x) + 16*A*m
^2*cos(e + f*x) - 8*C*m*cos(3*e + 3*f*x) + 4*C*m^2*cos(e + f*x) - 16*C*m*sin(2*e + 2*f*x)))/(2*f*(sin(e + f*x)
- 1)*(46*m + 36*m^2 + 8*m^3 + 15))

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