3.97 \(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=222 \[ \frac {a^2 (3 A-13 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {a^2 (3 A-13 B) \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}-\frac {a^2 (3 A-13 B) \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}} \]

[Out]

1/8*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(13/2)+1/48*a^2*(3*A-13*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(
9/2)-1/64*a^2*(3*A-13*B)*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(5/2)+1/256*a^2*(3*A-13*B)*cos(f*x+e)/c^3/f/(c-c*si
n(f*x+e))^(3/2)+1/512*a^2*(3*A-13*B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(9/2)/f*
2^(1/2)

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Rubi [A]  time = 0.51, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2967, 2859, 2680, 2650, 2649, 206} \[ \frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {a^2 (3 A-13 B) \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (3 A-13 B) \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (3 A-13 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^2*(3*A - 13*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(256*Sqrt[2]*c^(9/2)*f)
+ (a^2*(A + B)*c^2*Cos[e + f*x]^5)/(8*f*(c - c*Sin[e + f*x])^(13/2)) + (a^2*(3*A - 13*B)*Cos[e + f*x]^3)/(48*f
*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(3*A - 13*B)*Cos[e + f*x])/(64*c^2*f*(c - c*Sin[e + f*x])^(5/2)) + (a^2*(3
*A - 13*B)*Cos[e + f*x])/(256*c^3*f*(c - c*Sin[e + f*x])^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

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

Rule 2680

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

Rule 2859

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*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

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+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {1}{16} \left (a^2 (3 A-13 B) c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}}-\frac {\left (a^2 (3 A-13 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx}{32 c}\\ &=\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (3 A-13 B) \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {\left (a^2 (3 A-13 B)\right ) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{128 c^3}\\ &=\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (3 A-13 B) \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (3 A-13 B) \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (a^2 (3 A-13 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{512 c^4}\\ &=\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (3 A-13 B) \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (3 A-13 B) \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^2 (3 A-13 B)\right ) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{256 c^4 f}\\ &=\frac {a^2 (3 A-13 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{13/2}}+\frac {a^2 (3 A-13 B) \cos ^3(e+f x)}{48 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (3 A-13 B) \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 (3 A-13 B) \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 2.69, size = 357, normalized size = 1.61 \[ \frac {a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((-24-24 i) \sqrt [4]{-1} (3 A-13 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8+2013 A \sin \left (\frac {1}{2} (e+f x)\right )+999 A \sin \left (\frac {3}{2} (e+f x)\right )-69 A \sin \left (\frac {5}{2} (e+f x)\right )+9 A \sin \left (\frac {7}{2} (e+f x)\right )+2013 A \cos \left (\frac {1}{2} (e+f x)\right )-999 A \cos \left (\frac {3}{2} (e+f x)\right )-69 A \cos \left (\frac {5}{2} (e+f x)\right )-9 A \cos \left (\frac {7}{2} (e+f x)\right )+1517 B \sin \left (\frac {1}{2} (e+f x)\right )+791 B \sin \left (\frac {3}{2} (e+f x)\right )-725 B \sin \left (\frac {5}{2} (e+f x)\right )-39 B \sin \left (\frac {7}{2} (e+f x)\right )+1517 B \cos \left (\frac {1}{2} (e+f x)\right )-791 B \cos \left (\frac {3}{2} (e+f x)\right )-725 B \cos \left (\frac {5}{2} (e+f x)\right )+39 B \cos \left (\frac {7}{2} (e+f x)\right )\right )}{6144 f (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2*(2013*A*Cos[(e + f*x)/2] + 1517*B*Cos[(e + f*x
)/2] - 999*A*Cos[(3*(e + f*x))/2] - 791*B*Cos[(3*(e + f*x))/2] - 69*A*Cos[(5*(e + f*x))/2] - 725*B*Cos[(5*(e +
f*x))/2] - 9*A*Cos[(7*(e + f*x))/2] + 39*B*Cos[(7*(e + f*x))/2] - (24 + 24*I)*(-1)^(1/4)*(3*A - 13*B)*ArcTan[
(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8 + 2013*A*Sin[(e + f*x)/
2] + 1517*B*Sin[(e + f*x)/2] + 999*A*Sin[(3*(e + f*x))/2] + 791*B*Sin[(3*(e + f*x))/2] - 69*A*Sin[(5*(e + f*x)
)/2] - 725*B*Sin[(5*(e + f*x))/2] + 9*A*Sin[(7*(e + f*x))/2] - 39*B*Sin[(7*(e + f*x))/2]))/(6144*f*(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(9/2))

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fricas [B]  time = 0.49, size = 654, normalized size = 2.95 \[ -\frac {3 \, \sqrt {2} {\left ({\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{5} + 5 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - 20 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 8 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right ) + 16 \, {\left (3 \, A - 13 \, B\right )} a^{2} - {\left ({\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} - 4 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - 12 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 8 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right ) + 16 \, {\left (3 \, A - 13 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (3 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} + {\left (39 \, A + 343 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + 2 \, {\left (129 \, A + 209 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 12 \, {\left (13 \, A + 29 \, B\right )} a^{2} \cos \left (f x + e\right ) - 384 \, {\left (A + B\right )} a^{2} - {\left (3 \, {\left (3 \, A - 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} - 2 \, {\left (15 \, A + 191 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 12 \, {\left (19 \, A + 3 \, B\right )} a^{2} \cos \left (f x + e\right ) + 384 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3072 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f - {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/3072*(3*sqrt(2)*((3*A - 13*B)*a^2*cos(f*x + e)^5 + 5*(3*A - 13*B)*a^2*cos(f*x + e)^4 - 8*(3*A - 13*B)*a^2*c
os(f*x + e)^3 - 20*(3*A - 13*B)*a^2*cos(f*x + e)^2 + 8*(3*A - 13*B)*a^2*cos(f*x + e) + 16*(3*A - 13*B)*a^2 - (
(3*A - 13*B)*a^2*cos(f*x + e)^4 - 4*(3*A - 13*B)*a^2*cos(f*x + e)^3 - 12*(3*A - 13*B)*a^2*cos(f*x + e)^2 + 8*(
3*A - 13*B)*a^2*cos(f*x + e) + 16*(3*A - 13*B)*a^2)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 - 2*sqrt(2)*s
qrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)
*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(3*(3*A - 13*B
)*a^2*cos(f*x + e)^4 + (39*A + 343*B)*a^2*cos(f*x + e)^3 + 2*(129*A + 209*B)*a^2*cos(f*x + e)^2 - 12*(13*A + 2
9*B)*a^2*cos(f*x + e) - 384*(A + B)*a^2 - (3*(3*A - 13*B)*a^2*cos(f*x + e)^3 - 2*(15*A + 191*B)*a^2*cos(f*x +
e)^2 + 12*(19*A + 3*B)*a^2*cos(f*x + e) + 384*(A + B)*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^5*f*cos
(f*x + e)^5 + 5*c^5*f*cos(f*x + e)^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e)
+ 16*c^5*f - (c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e)
+ 16*c^5*f)*sin(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*2*(1/1536*(-1527*A*a^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(
c*tan((f*x+exp(1))/2)^2+c))^15-39*B*a^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^15-4473
*A*a^2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^14+2487*B*a^2*sqrt(c)*(-sqrt(c)*
tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^14-22233*A*a^2*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan
((f*x+exp(1))/2)^2+c))^13-7593*B*a^2*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^13-23811
*A*a^2*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^12+1293*B*a^2*sqrt(c)*c*(-sqrt
(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^12+2133*A*a^2*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(
c*tan((f*x+exp(1))/2)^2+c))^11-1563*B*a^2*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^1
1+68019*A*a^2*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^10-10589*B*a^2*sqrt(c
)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^10+25371*A*a^2*c^3*(-sqrt(c)*tan((f*x+exp
(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9+9355*B*a^2*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1)
)/2)^2+c))^9-71487*A*a^2*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-3055*B*a
^2*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-25173*A*a^2*c^4*(-sqrt(c)*tan(
(f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7+7195*B*a^2*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*
x+exp(1))/2)^2+c))^7+56469*A*a^2*sqrt(c)*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6+
15909*B*a^2*sqrt(c)*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-10971*A*a^2*c^5*(-sqr
t(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5-2123*B*a^2*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(
c*tan((f*x+exp(1))/2)^2+c))^5-31881*A*a^2*sqrt(c)*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)
^2+c))^4-3673*B*a^2*sqrt(c)*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+17079*A*a^2*c
^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-5913*B*a^2*c^6*(-sqrt(c)*tan((f*x+exp(1))/
2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+345*A*a^2*c^7*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2
+c))-7695*A*a^2*sqrt(c)*c^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+41*B*a^2*c^7*(-sq
rt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-3007*B*a^2*sqrt(c)*c^6*(-sqrt(c)*tan((f*x+exp(1))/2
)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-117*A*a^2*sqrt(c)*c^7-5*B*a^2*sqrt(c)*c^7)/c^4/(-(-sqrt(c)*tan((f*x+exp(1
))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+
c))+c)^8/sign(tan((f*x+exp(1))/2)-1)+1/512*(3*A*a^2-13*B*a^2)*atan((-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(
c*tan((f*x+exp(1))/2)^2+c))/sqrt(2)/sqrt(-c))/sqrt(2)/c^4/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1))

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maple [B]  time = 1.70, size = 440, normalized size = 1.98 \[ \frac {a^{2} \left (-12 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{5} \left (3 A -13 B \right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+24 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{5} \left (3 A -13 B \right ) \sin \left (f x +e \right )-3 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{5} \left (3 A -13 B \right ) \left (\cos ^{4}\left (f x +e \right )\right )+24 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{5} \left (3 A -13 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+144 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {9}{2}}-264 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {7}{2}}-132 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{\frac {5}{2}}+18 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {7}{2}} c^{\frac {3}{2}}-624 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {9}{2}}+1144 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {7}{2}}-452 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{\frac {5}{2}}-78 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {7}{2}} c^{\frac {3}{2}}-72 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}+312 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{1536 c^{\frac {19}{2}} \left (\sin \left (f x +e \right )-1\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x)

[Out]

1/1536/c^(19/2)*a^2*(-12*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^5*(3*A-13*B)*sin(f*x+e)
*cos(f*x+e)^2+24*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^5*(3*A-13*B)*sin(f*x+e)-3*arcta
nh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^5*(3*A-13*B)*cos(f*x+e)^4+24*arctanh(1/2*(c+c*sin(f*x
+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^5*(3*A-13*B)*cos(f*x+e)^2+144*A*(c+c*sin(f*x+e))^(1/2)*c^(9/2)-264*A*(c+
c*sin(f*x+e))^(3/2)*c^(7/2)-132*A*(c+c*sin(f*x+e))^(5/2)*c^(5/2)+18*A*(c+c*sin(f*x+e))^(7/2)*c^(3/2)-624*B*(c+
c*sin(f*x+e))^(1/2)*c^(9/2)+1144*B*(c+c*sin(f*x+e))^(3/2)*c^(7/2)-452*B*(c+c*sin(f*x+e))^(5/2)*c^(5/2)-78*B*(c
+c*sin(f*x+e))^(7/2)*c^(3/2)-72*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^5+312*B*2^(1/2
)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^5)*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)^3/cos(f*x+e
)/(c-c*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(9/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(9/2), 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+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

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