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

Optimal. Leaf size=118 \[ -\frac {4 c^2 (3 A-5 B) \cos (e+f x)}{3 a f \sqrt {c-c \sin (e+f x)}}-\frac {c (3 A-5 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f} \]

[Out]

-(A-B)*sec(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/c/f-4/3*(3*A-5*B)*c^2*cos(f*x+e)/a/f/(c-c*sin(f*x+e))^(1/2)-1/3*(3*
A-5*B)*c*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f

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Rubi [A]  time = 0.32, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2967, 2855, 2647, 2646} \[ -\frac {4 c^2 (3 A-5 B) \cos (e+f x)}{3 a f \sqrt {c-c \sin (e+f x)}}-\frac {c (3 A-5 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(3*A - 5*B)*c^2*Cos[e + f*x])/(3*a*f*Sqrt[c - c*Sin[e + f*x]]) - ((3*A - 5*B)*c*Cos[e + f*x]*Sqrt[c - c*Si
n[e + f*x]])/(3*a*f) - ((A - B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(a*c*f)

Rule 2646

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

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2967

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^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), 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] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 113, normalized size = 0.96 \[ \frac {c \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) ((14 B-6 A) \sin (e+f x)-18 A+B \cos (2 (e+f x))+27 B)}{3 a f (\sin (e+f x)+1) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-18*A + 27*B + B*Cos[2*(e + f*x)] + (-6*A + 14*B)*Sin[e + f*x])*Sqrt
[c - c*Sin[e + f*x]])/(3*a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))

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fricas [A]  time = 0.43, size = 67, normalized size = 0.57 \[ \frac {2 \, {\left (B c \cos \left (f x + e\right )^{2} - {\left (3 \, A - 7 \, B\right )} c \sin \left (f x + e\right ) - {\left (9 \, A - 13 \, B\right )} c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(B*c*cos(f*x + e)^2 - (3*A - 7*B)*c*sin(f*x + e) - (9*A - 13*B)*c)*sqrt(-c*sin(f*x + e) + c)/(a*f*cos(f*x
+ e))

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

[Out]

Timed out

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maple [A]  time = 1.12, size = 73, normalized size = 0.62 \[ \frac {2 c^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right ) \left (3 A -7 B \right )-B \left (\cos ^{2}\left (f x +e \right )\right )+9 A -13 B \right )}{3 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*c^2/a*(sin(f*x+e)-1)*(sin(f*x+e)*(3*A-7*B)-B*cos(f*x+e)^2+9*A-13*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [B]  time = 0.46, size = 294, normalized size = 2.49 \[ \frac {2 \, {\left (\frac {3 \, {\left (3 \, c^{\frac {3}{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} A}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (7 \, c^{\frac {3}{2}} + \frac {7 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {12 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {7 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {7 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} B}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*(3*c^(3/2) + 2*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 6*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 2*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*A/((a + a*sin
(f*x + e)/(cos(f*x + e) + 1))*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)) - 2*(7*c^(3/2) + 7*c^(3/2)*sin(
f*x + e)/(cos(f*x + e) + 1) + 12*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*c^(3/2)*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 7*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*B/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*(si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}}{a+a\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A c \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {A c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {B c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {B c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\right )\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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