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]
(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)
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Rubi [A] time = 0.479186, antiderivative size = 154, normalized size of antiderivative = 1.,
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 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]))
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 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 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]
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*}
Mathematica [A] time = 1.21923, 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]
-(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)]))/(6*a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 +
Sin[e + f*x])^2)
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Maple [A] time = 0.905, size = 105, normalized size = 0.7 \begin{align*} -{\frac{2\,{c}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( -B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( 18\,A-50\,B \right ) \sin \left ( fx+e \right ) + \left ( -3\,A+12\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+14\,A-46\,B \right ) }{3\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}
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*(-1+sin(f*x+e))/(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 = 1.5569, size = 779, normalized size = 5.06 \begin{align*} \text{result too large to display} \end{align*}
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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Fricas [A] time = 1.49868, size = 270, normalized size = 1.75 \begin{align*} -\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 )}} \end{align*}
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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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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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Giac [B] time = 2.16073, size = 1339, normalized size = 8.69 \begin{align*} \text{result too large to display} \end{align*}
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]
-1/3*((21*sqrt(2)*A*a^8*sqrt(c) - 91*sqrt(2)*B*a^8*sqrt(c) - 30*A*a^8*sqrt(c) + 130*B*a^8*sqrt(c) - 10*sqrt(2)
*A*c^(13/2) + 46*sqrt(2)*B*c^(13/2) + 12*A*c^(13/2) - 60*B*c^(13/2))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(5*sqrt(2)*
a^2*c^4 - 7*a^2*c^4) + ((((3*A*a^6*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 14*B*a^6*c^4*sgn(tan(1/2*f*x + 1/2*e) -
1))*tan(1/2*f*x + 1/2*e)/c^6 + 3*(A*a^6*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 4*B*a^6*c^4*sgn(tan(1/2*f*x + 1/2
*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + 3*(A*a^6*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 4*B*a^6*c^4*sgn(tan(1/2*f*x
+ 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + (3*A*a^6*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 14*B*a^6*c^4*sgn(tan(
1/2*f*x + 1/2*e) - 1))/c^6)/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(3/2) - 16*(3*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(
c*tan(1/2*f*x + 1/2*e)^2 + c))^5*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 6*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(
c*tan(1/2*f*x + 1/2*e)^2 + c))^4*A*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) -
sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*B*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 4*(sqrt(c)*tan(1/2*f*x + 1/2*e
) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 10*(sqrt(c)*tan(1/2*f*x + 1/2*
e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 12*(sqrt(c)*tan(1/2*f*x + 1/2
*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 30*(sqrt(c)*tan(1/2*f*x
+ 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 12*(sqrt(c)*tan(1/2
*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 27*(sqrt(c)*tan(1/2*
f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 2*A*c^(11/2)*sgn(tan(
1/2*f*x + 1/2*e) - 1) - 5*B*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*t
an(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c)
- c)^3*a^2))/f