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

Optimal. Leaf size=154 \[ -\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f}+\frac {32 c^2 (A-3 B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f}-\frac {(A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f}-\frac {8 c (A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f} \]

[Out]

-8/3*(A-3*B)*c*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f-1/3*(A-3*B)*sec(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a^2/f-1/3
*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(9/2)/a^2/c^2/f+32/3*(A-3*B)*c^2*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f

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Rubi [A]  time = 0.48, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2967, 2855, 2674, 2673} \[ -\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f}+\frac {32 c^2 (A-3 B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f}-\frac {(A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f}-\frac {8 c (A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^2,x]

[Out]

(32*(A - 3*B)*c^2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*a^2*f) - (8*(A - 3*B)*c*Sec[e + f*x]*(c - c*Sin[e
+ f*x])^(3/2))/(3*a^2*f) - ((A - 3*B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*a^2*f) - ((A - B)*Sec[e + f*
x]^3*(c - c*Sin[e + f*x])^(9/2))/(3*a^2*c^2*f)

Rule 2673

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

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 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))^{5/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac {\int \sec ^4(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2} \, dx}{a^2 c^2}\\ &=-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f}-\frac {(A-3 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{2 a^2 c}\\ &=-\frac {(A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f}-\frac {(4 (A-3 B)) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{3 a^2}\\ &=-\frac {8 (A-3 B) c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {(A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f}-\frac {(16 (A-3 B) c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{3 a^2}\\ &=\frac {32 (A-3 B) c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f}-\frac {8 (A-3 B) c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {(A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 1.16, size = 130, normalized size = 0.84 \[ -\frac {c^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) ((201 B-72 A) \sin (e+f x)+6 (A-4 B) \cos (2 (e+f x))-50 A+B \sin (3 (e+f x))+160 B)}{6 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^2,x]

[Out]

-1/6*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-50*A + 160*B + 6*(A - 4*B)*Cos[2*(e
+ f*x)] + (-72*A + 201*B)*Sin[e + f*x] + B*Sin[3*(e + f*x)]))/(a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1
+ Sin[e + f*x])^2)

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fricas [A]  time = 0.44, size = 110, normalized size = 0.71 \[ -\frac {2 \, {\left (3 \, {\left (A - 4 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (7 \, A - 23 \, B\right )} c^{2} + {\left (B c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (9 \, A - 25 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-2/3*(3*(A - 4*B)*c^2*cos(f*x + e)^2 - 2*(7*A - 23*B)*c^2 + (B*c^2*cos(f*x + e)^2 - 2*(9*A - 25*B)*c^2)*sin(f*
x + e))*sqrt(-c*sin(f*x + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*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))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.27, size = 105, normalized size = 0.68 \[ -\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (18 A -50 B \right ) \sin \left (f x +e \right )+\left (-3 A +12 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+14 A -46 B \right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x)

[Out]

-2/3*c^3/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(-B*cos(f*x+e)^2*sin(f*x+e)+(18*A-50*B)*sin(f*x+e)+(-3*A+12*B)*cos(
f*x+e)^2+14*A-46*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [B]  time = 0.46, size = 577, normalized size = 3.75 \[ -\frac {2 \, {\left (\frac {{\left (11 \, c^{\frac {5}{2}} + \frac {36 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {56 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {108 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {90 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {108 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {56 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {36 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {11 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} A}{{\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} - \frac {2 \, {\left (17 \, c^{\frac {5}{2}} + \frac {51 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {92 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {149 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {150 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {149 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {92 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {51 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {17 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} B}{{\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{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))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*((11*c^(5/2) + 36*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 56*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 + 108*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 90*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 108*c^(
5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 56*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 36*c^(5/2)*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7 + 11*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*A/((a^2 + 3*a^2*sin(f*x + e)/(c
os(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)) - 2*(17*c^(5/2) + 51*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 92*c
^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 149*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 150*c^(5/2)*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 + 149*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 92*c^(5/2)*sin(f*x + e)^6
/(cos(f*x + e) + 1)^6 + 51*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 17*c^(5/2)*sin(f*x + e)^8/(cos(f*x +
e) + 1)^8)*B/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*s
in(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x))^2,x)

[Out]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x))^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+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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