3.106 \(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=217 \[ -\frac {5 a^3 (A-15 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}+\frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 c (A-15 B) \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/8*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(15/2)+1/48*a^3*(A-15*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(
11/2)-5/192*a^3*(A-15*B)*cos(f*x+e)^3/c/f/(c-c*sin(f*x+e))^(7/2)+5/128*a^3*(A-15*B)*cos(f*x+e)/c^3/f/(c-c*sin(
f*x+e))^(3/2)-5/256*a^3*(A-15*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 = 0.56, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2967, 2859, 2680, 2649, 206} \[ \frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (A-15 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}+\frac {a^3 c (A-15 B) \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(-5*a^3*(A - 15*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*c^(9/2)*f)
+ (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(8*f*(c - c*Sin[e + f*x])^(15/2)) + (a^3*(A - 15*B)*c*Cos[e + f*x]^5)/(48*
f*(c - c*Sin[e + f*x])^(11/2)) - (5*a^3*(A - 15*B)*Cos[e + f*x]^3)/(192*c*f*(c - c*Sin[e + f*x])^(7/2)) + (5*a
^3*(A - 15*B)*Cos[e + f*x])/(128*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 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))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{15/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {1}{16} \left (a^3 (A-15 B) c^2\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 (A-15 B) c \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {1}{96} \left (5 a^3 (A-15 B)\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 (A-15 B) c \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}}+\frac {\left (5 a^3 (A-15 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{128 c^2}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 (A-15 B) c \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (5 a^3 (A-15 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{256 c^4}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 (A-15 B) c \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (5 a^3 (A-15 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 )}{128 c^4 f}\\ &=-\frac {5 a^3 (A-15 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 (A-15 B) c \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 4.38, size = 355, normalized size = 1.64 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((120+120 i) \sqrt [4]{-1} (A-15 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8+1765 A \sin \left (\frac {1}{2} (e+f x)\right )+895 A \sin \left (\frac {3}{2} (e+f x)\right )-397 A \sin \left (\frac {5}{2} (e+f x)\right )-15 A \sin \left (\frac {7}{2} (e+f x)\right )+1765 A \cos \left (\frac {1}{2} (e+f x)\right )-895 A \cos \left (\frac {3}{2} (e+f x)\right )-397 A \cos \left (\frac {5}{2} (e+f x)\right )+15 A \cos \left (\frac {7}{2} (e+f x)\right )+405 B \sin \left (\frac {1}{2} (e+f x)\right )+2703 B \sin \left (\frac {3}{2} (e+f x)\right )+579 B \sin \left (\frac {5}{2} (e+f x)\right )-543 B \sin \left (\frac {7}{2} (e+f x)\right )+405 B \cos \left (\frac {1}{2} (e+f x)\right )-2703 B \cos \left (\frac {3}{2} (e+f x)\right )+579 B \cos \left (\frac {5}{2} (e+f x)\right )+543 B \cos \left (\frac {7}{2} (e+f x)\right )\right )}{3072 f (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(1765*A*Cos[(e + f*x)/2] + 405*B*Cos[(e + f*x)
/2] - 895*A*Cos[(3*(e + f*x))/2] - 2703*B*Cos[(3*(e + f*x))/2] - 397*A*Cos[(5*(e + f*x))/2] + 579*B*Cos[(5*(e
+ f*x))/2] + 15*A*Cos[(7*(e + f*x))/2] + 543*B*Cos[(7*(e + f*x))/2] + (120 + 120*I)*(-1)^(1/4)*(A - 15*B)*ArcT
an[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8 + 1765*A*Sin[(e + f*
x)/2] + 405*B*Sin[(e + f*x)/2] + 895*A*Sin[(3*(e + f*x))/2] + 2703*B*Sin[(3*(e + f*x))/2] - 397*A*Sin[(5*(e +
f*x))/2] + 579*B*Sin[(5*(e + f*x))/2] - 15*A*Sin[(7*(e + f*x))/2] - 543*B*Sin[(7*(e + f*x))/2]))/(3072*f*(Cos[
(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(9/2))

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fricas [B]  time = 0.48, size = 633, normalized size = 2.92 \[ -\frac {15 \, \sqrt {2} {\left ({\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{5} + 5 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - 8 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 20 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 8 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right ) + 16 \, {\left (A - 15 \, B\right )} a^{3} - {\left ({\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - 4 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 12 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 8 \, {\left (A - 15 \, B\right )} a^{3} \cos \left (f x + e\right ) + 16 \, {\left (A - 15 \, B\right )} a^{3}\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 (5 \, A + 181 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - {\left (191 \, A - 561 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 2 \, {\left (169 \, A + 537 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 12 \, {\left (21 \, A - 59 \, B\right )} a^{3} \cos \left (f x + e\right ) + 384 \, {\left (A + B\right )} a^{3} - {\left (3 \, {\left (5 \, A + 181 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 2 \, {\left (103 \, A - 9 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 12 \, {\left (11 \, A + 91 \, B\right )} a^{3} \cos \left (f x + e\right ) - 384 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1536 \, {\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))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/1536*(15*sqrt(2)*((A - 15*B)*a^3*cos(f*x + e)^5 + 5*(A - 15*B)*a^3*cos(f*x + e)^4 - 8*(A - 15*B)*a^3*cos(f*
x + e)^3 - 20*(A - 15*B)*a^3*cos(f*x + e)^2 + 8*(A - 15*B)*a^3*cos(f*x + e) + 16*(A - 15*B)*a^3 - ((A - 15*B)*
a^3*cos(f*x + e)^4 - 4*(A - 15*B)*a^3*cos(f*x + e)^3 - 12*(A - 15*B)*a^3*cos(f*x + e)^2 + 8*(A - 15*B)*a^3*cos
(f*x + e) + 16*(A - 15*B)*a^3)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-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*(5*A + 181*B)*a^3*cos(f*x + e)^4
- (191*A - 561*B)*a^3*cos(f*x + e)^3 - 2*(169*A + 537*B)*a^3*cos(f*x + e)^2 + 12*(21*A - 59*B)*a^3*cos(f*x +
e) + 384*(A + B)*a^3 - (3*(5*A + 181*B)*a^3*cos(f*x + e)^3 + 2*(103*A - 9*B)*a^3*cos(f*x + e)^2 - 12*(11*A + 9
1*B)*a^3*cos(f*x + e) - 384*(A + B)*a^3)*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))^3*(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*(1/768*(-783*A*a^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*
tan((f*x+exp(1))/2)^2+c))^15+225*B*a^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^15-993*A
*a^3*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^14+4911*B*a^3*sqrt(c)*(-sqrt(c)*ta
n((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^14-14913*A*a^3*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((
f*x+exp(1))/2)^2+c))^13+14031*B*a^3*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^13-11259*
A*a^3*sqrt(c)*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^12+77493*B*a^3*sqrt(c)*c*(-sqrt
(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^12+285*A*a^3*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c
*tan((f*x+exp(1))/2)^2+c))^11+54861*B*a^3*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^1
1+28715*A*a^3*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^10-124293*B*a^3*sqrt(
c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^10+17363*A*a^3*c^3*(-sqrt(c)*tan((f*x+ex
p(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9-73821*B*a^3*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(
1))/2)^2+c))^9-37271*A*a^3*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8+89817*
B*a^3*sqrt(c)*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-8989*A*a^3*c^4*(-sqrt(c)*ta
n((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7-10317*B*a^3*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan(
(f*x+exp(1))/2)^2+c))^7+36189*A*a^3*sqrt(c)*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))
^6-32115*B*a^3*sqrt(c)*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-6547*A*a^3*c^5*(-s
qrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5+71325*B*a^3*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sq
rt(c*tan((f*x+exp(1))/2)^2+c))^5-17777*A*a^3*sqrt(c)*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))
/2)^2+c))^4-7521*B*a^3*sqrt(c)*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+5583*A*a^3
*c^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3-35361*B*a^3*c^6*(-sqrt(c)*tan((f*x+exp(1
))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+193*A*a^3*c^7*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2
)^2+c))-5351*A*a^3*sqrt(c)*c^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-2127*B*a^3*c^7
*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))+10377*B*a^3*sqrt(c)*c^6*(-sqrt(c)*tan((f*x+exp
(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-61*A*a^3*sqrt(c)*c^7+147*B*a^3*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)+1/256*(-5*A*a^3+75*B*a^3)*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 = 1.88, size = 432, normalized size = 1.99 \[ \frac {a^{3} \left (60 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} \left (A -15 B \right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-120 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} \left (A -15 B \right ) \sin \left (f x +e \right )+15 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} \left (A -15 B \right ) \left (\cos ^{4}\left (f x +e \right )\right )-120 \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4} \left (A -15 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )-30 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {7}{2}} \sqrt {c}-292 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{\frac {3}{2}}+440 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {5}{2}}-240 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {7}{2}}-1086 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {7}{2}} \sqrt {c}+4380 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{\frac {3}{2}}-6600 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {5}{2}}+3600 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {7}{2}}+120 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}-1800 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{768 c^{\frac {17}{2}} \left (\sin \left (f x +e \right )-1\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x)

[Out]

1/768*a^3*(60*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^4*(A-15*B)*sin(f*x+e)*cos(f*x+e)^2
-120*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^4*(A-15*B)*sin(f*x+e)+15*arctanh(1/2*(c+c*s
in(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^4*(A-15*B)*cos(f*x+e)^4-120*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(
1/2)/c^(1/2))*2^(1/2)*c^4*(A-15*B)*cos(f*x+e)^2-30*A*(c+c*sin(f*x+e))^(7/2)*c^(1/2)-292*A*(c+c*sin(f*x+e))^(5/
2)*c^(3/2)+440*A*(c+c*sin(f*x+e))^(3/2)*c^(5/2)-240*A*(c+c*sin(f*x+e))^(1/2)*c^(7/2)-1086*B*(c+c*sin(f*x+e))^(
7/2)*c^(1/2)+4380*B*(c+c*sin(f*x+e))^(5/2)*c^(3/2)-6600*B*(c+c*sin(f*x+e))^(3/2)*c^(5/2)+3600*B*(c+c*sin(f*x+e
))^(1/2)*c^(7/2)+120*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4-1800*B*2^(1/2)*arctanh(
1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4)*(c*(1+sin(f*x+e)))^(1/2)/c^(17/2)/(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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(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)^3/(-c*sin(f*x + e) + c)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^(9/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(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))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

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