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 \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac{5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac{5 a^3 (A-15 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} c^{9/2} f}+\frac{a^3 c (A-15 B) \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac{5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}} \]
[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))
________________________________________________________________________________________
Rubi [A] time = 0.557293, antiderivative size = 217, normalized size of antiderivative = 1.,
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 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 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 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 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 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])
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*}
Mathematica [C] time = 4.67189, 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))
________________________________________________________________________________________
Maple [B] time = 1.585, size = 432, normalized size = 2. \begin{align*}{\frac{{a}^{3}}{768\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) f} \left ( 60\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{4} \left ( A-15\,B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-120\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{4} \left ( A-15\,B \right ) \sin \left ( fx+e \right ) +15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{4} \left ( A-15\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-120\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{c}^{4} \left ( A-15\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-240\,A\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{7/2}+440\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}{c}^{5/2}-292\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{3/2}-30\,A \left ( c+c\sin \left ( fx+e \right ) \right ) ^{7/2}\sqrt{c}+3600\,B\sqrt{c+c\sin \left ( fx+e \right ) }{c}^{7/2}-6600\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}{c}^{5/2}+4380\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{3/2}-1086\,B \left ( c+c\sin \left ( fx+e \right ) \right ) ^{7/2}\sqrt{c}+120\,A\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4}-1800\,B\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{4} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{17}{2}}}{\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+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)^2-240*A*(c+c*sin(f*x+e))^(1/2)*c^(7/2)+440*A*(c+c*sin(f*x+e))^(3
/2)*c^(5/2)-292*A*(c+c*sin(f*x+e))^(5/2)*c^(3/2)-30*A*(c+c*sin(f*x+e))^(7/2)*c^(1/2)+3600*B*(c+c*sin(f*x+e))^(
1/2)*c^(7/2)-6600*B*(c+c*sin(f*x+e))^(3/2)*c^(5/2)+4380*B*(c+c*sin(f*x+e))^(5/2)*c^(3/2)-1086*B*(c+c*sin(f*x+e
))^(7/2)*c^(1/2)+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)/(-1+sin(f*x+e))^3/cos(f*x+e
)/(c-c*sin(f*x+e))^(1/2)/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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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Fricas [B] time = 1.63184, size = 1646, normalized size = 7.59 \begin{align*} \text{result too large to display} \end{align*}
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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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+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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Giac [B] time = 7.61863, size = 2140, normalized size = 9.86 \begin{align*} \text{result too large to display} \end{align*}
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]
-1/384*(15*sqrt(2)*(A*a^3 - 15*B*a^3)*arctan(-1/2*sqrt(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))/sqrt(-c))/(sqrt(-c)*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1)) - 2*(783*(sqrt(c)*tan(1/2*f*x
+ 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^15*A*a^3 - 225*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*
f*x + 1/2*e)^2 + c))^15*B*a^3 - 993*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^14*A*a
^3*sqrt(c) + 4911*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^14*B*a^3*sqrt(c) + 14913
*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^13*A*a^3*c - 14031*(sqrt(c)*tan(1/2*f*x +
1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^13*B*a^3*c - 11259*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/
2*f*x + 1/2*e)^2 + c))^12*A*a^3*c^(3/2) + 77493*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c))^12*B*a^3*c^(3/2) - 285*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^11*A*a^3*c^2
- 54861*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^11*B*a^3*c^2 + 28715*(sqrt(c)*tan(
1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^10*A*a^3*c^(5/2) - 124293*(sqrt(c)*tan(1/2*f*x + 1/2*e)
- sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^10*B*a^3*c^(5/2) - 17363*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2
*f*x + 1/2*e)^2 + c))^9*A*a^3*c^3 + 73821*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^
9*B*a^3*c^3 - 37271*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*A*a^3*c^(7/2) + 8981
7*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*B*a^3*c^(7/2) + 8989*(sqrt(c)*tan(1/2*
f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*A*a^3*c^4 + 10317*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*
tan(1/2*f*x + 1/2*e)^2 + c))^7*B*a^3*c^4 + 36189*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c))^6*A*a^3*c^(9/2) - 32115*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*B*a^3*c^(
9/2) + 6547*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*A*a^3*c^5 - 71325*(sqrt(c)*t
an(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*B*a^3*c^5 - 17777*(sqrt(c)*tan(1/2*f*x + 1/2*e) -
sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*A*a^3*c^(11/2) - 7521*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x
+ 1/2*e)^2 + c))^4*B*a^3*c^(11/2) - 5583*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^
3*A*a^3*c^6 + 35361*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*a^3*c^6 - 5351*(sq
rt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*a^3*c^(13/2) + 10377*(sqrt(c)*tan(1/2*f*x
+ 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*a^3*c^(13/2) - 193*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*
tan(1/2*f*x + 1/2*e)^2 + c))*A*a^3*c^7 + 2127*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 +
c))*B*a^3*c^7 - 61*A*a^3*c^(15/2) + 147*B*a^3*c^(15/2))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(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)^8*c^4*s
gn(tan(1/2*f*x + 1/2*e) - 1)))/f