Integrand size = 38, antiderivative size = 225 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {5 a^3 (3 A+11 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}} \]
1/4*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(11/2)-1/8*a^3*(3*A+11*B )*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(7/2)-5/24*a^3*(3*A+11*B)*cos(f*x+e)^3 /c/f/(c-c*sin(f*x+e))^(3/2)+5/4*a^3*(3*A+11*B)*arctanh(1/2*cos(f*x+e)*c^(1 /2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)-5/4*a^3*(3*A+11*B)*c os(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 12.11 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.93 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-3 (9 A+17 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(15+15 i) \sqrt [4]{-1} (3 A+11 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-6 (2 A+11 B) \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+2 B \cos \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-6 (9 A+17 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )-6 (2 A+11 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )-2 B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {3}{2} (e+f x)\right )\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^{5/2}} \]
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(12*(A + B )*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 3*(9*A + 17*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (15 + 15*I)*(-1)^(1/4)*(3*A + 11*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])^4 - 6*(2*A + 11*B)*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x) /2])^4 + 2*B*Cos[(3*(e + f*x))/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 24*(A + B)*Sin[(e + f*x)/2] - 6*(9*A + 17*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] - 6*(2*A + 11*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 2*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^ 4*Sin[(3*(e + f*x))/2]))/(6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(5/2))
Time = 1.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3446, 3042, 3338, 3042, 3159, 3042, 3158, 3042, 3158, 3042, 3128, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{9/2}}dx}{8 c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{5/2}}dx}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \left (\frac {2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \left (\frac {2 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \left (\frac {2 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {(3 A+11 B) \left (\frac {\cos ^5(e+f x)}{c f (c-c \sin (e+f x))^{7/2}}-\frac {5 \left (\frac {2 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{2 c^2}\right )}{8 c}\right )\) |
a^3*c^3*(((A + B)*Cos[e + f*x]^7)/(4*f*(c - c*Sin[e + f*x])^(11/2)) - ((3* A + 11*B)*(Cos[e + f*x]^5/(c*f*(c - c*Sin[e + f*x])^(7/2)) - (5*((-2*Cos[e + f*x]^3)/(3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (2*((2*Sqrt[2]*ArcTanh[(Sq rt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) - (2* Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])))/c))/(2*c^2)))/(8*c))
3.2.4.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
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] + Simp[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] && EqQ[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]
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] + Simp[(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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[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] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(433\) vs. \(2(198)=396\).
Time = 3.98 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.93
| method | result | size |
| default | \(-\frac {a^{3} \left (\left (24 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+8 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}+120 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-165 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+\sin \left (f x +e \right ) \left (48 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-90 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+16 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}+240 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-330 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right )+54 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}-132 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+90 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+86 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}-420 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+330 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{12 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(434\) |
| parts | \(\text {Expression too large to display}\) | \(1117\) |
-1/12/c^(9/2)*a^3*((24*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-45*A*2^(1/2)*arcta nh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+8*B*(c+c*sin(f*x+e))^(3 /2)*c^(1/2)+120*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-165*B*2^(1/2)*arctanh(1/2 *(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2)*cos(f*x+e)^2+sin(f*x+e)*(48* A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-90*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e)) ^(1/2)*2^(1/2)/c^(1/2))*c^2+16*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2)+240*B*(c+c *sin(f*x+e))^(1/2)*c^(3/2)-330*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2 )*2^(1/2)/c^(1/2))*c^2)+54*A*(c+c*sin(f*x+e))^(3/2)*c^(1/2)-132*A*(c+c*sin (f*x+e))^(1/2)*c^(3/2)+90*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^( 1/2)/c^(1/2))*c^2+86*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2)-420*B*(c+c*sin(f*x+e ))^(1/2)*c^(3/2)+330*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/ c^(1/2))*c^2)*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin( f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (198) = 396\).
Time = 0.29 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.24 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {15 \, \sqrt {2} {\left ({\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right ) - 4 \, {\left (3 \, A + 11 \, B\right )} a^{3} - {\left ({\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right ) - 4 \, {\left (3 \, A + 11 \, 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 (4 \, B a^{3} \cos \left (f x + e\right )^{4} - 4 \, {\left (3 \, A + 14 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 3 \, {\left (13 \, A + 37 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (13 \, A + 53 \, B\right )} a^{3} \cos \left (f x + e\right ) - 12 \, {\left (A + B\right )} a^{3} - {\left (4 \, B a^{3} \cos \left (f x + e\right )^{3} + 12 \, {\left (A + 5 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (17 \, A + 57 \, B\right )} a^{3} \cos \left (f x + e\right ) + 12 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{24 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
1/24*(15*sqrt(2)*((3*A + 11*B)*a^3*cos(f*x + e)^3 + 3*(3*A + 11*B)*a^3*cos (f*x + e)^2 - 2*(3*A + 11*B)*a^3*cos(f*x + e) - 4*(3*A + 11*B)*a^3 - ((3*A + 11*B)*a^3*cos(f*x + e)^2 - 2*(3*A + 11*B)*a^3*cos(f*x + e) - 4*(3*A + 1 1*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*(4*B*a^3*cos(f*x + e)^ 4 - 4*(3*A + 14*B)*a^3*cos(f*x + e)^3 + 3*(13*A + 37*B)*a^3*cos(f*x + e)^2 + 3*(13*A + 53*B)*a^3*cos(f*x + e) - 12*(A + B)*a^3 - (4*B*a^3*cos(f*x + e)^3 + 12*(A + 5*B)*a^3*cos(f*x + e)^2 + 3*(17*A + 57*B)*a^3*cos(f*x + e) + 12*(A + B)*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f* cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))
Timed out. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 778 vs. \(2 (198) = 396\).
Time = 0.45 (sec) , antiderivative size = 778, normalized size of antiderivative = 3.46 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
1/96*(60*sqrt(2)*(3*A*a^3*sqrt(c) + 11*B*a^3*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(c^3*sgn(sin(- 1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(A*a^3*sqrt(c) + B*a^3*sqrt(c) + 1 6*A*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f* x + 1/2*e) + 1) + 32*B*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(c os(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 90*A*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f *x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 330*B*a^3*sqrt (c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2/(c^3*(cos(-1/4*pi + 1/2*f* x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 3*(16*sqrt(2)*A*a ^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 32*sqrt(2)*B*a^3*c^(7/2)* (cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/( cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + sqrt(2)*A*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + sqrt(2)*B*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/c^6 - 128*sqrt(2)*(3*A*a^3*sqrt(c) + 17*B*a^3*sqrt(c) - 6 *A*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 30*B*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/...
Timed out. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]