Integrand size = 37, antiderivative size = 308 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=-\frac {a^{5/2} \left (A d \left (3 c^2+10 c d+19 d^2\right )-B \left (15 c^3+30 c^2 d+7 c d^2-20 d^3\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{7/2} (c+d)^{5/2} f}+\frac {a^3 \left (3 A d (c+3 d)-B \left (15 c^2+25 c d+4 d^2\right )\right ) \cos (e+f x)}{4 d^3 (c+d)^2 f \sqrt {a+a \sin (e+f x)}}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{4 d^2 (c+d)^2 f (c+d \sin (e+f x))} \]
-1/4*a^(5/2)*(A*d*(3*c^2+10*c*d+19*d^2)-B*(15*c^3+30*c^2*d+7*c*d^2-20*d^3) )*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/d ^(7/2)/(c+d)^(5/2)/f+1/2*a*(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/d/ (c+d)/f/(c+d*sin(f*x+e))^2+1/4*a^3*(3*A*d*(c+3*d)-B*(15*c^2+25*c*d+4*d^2)) *cos(f*x+e)/d^3/(c+d)^2/f/(a+a*sin(f*x+e))^(1/2)-1/4*a^2*(A*d*(c+7*d)-B*(5 *c^2+7*c*d-4*d^2))*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d^2/(c+d)^2/f/(c+d*si n(f*x+e))
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 12.86 (sec) , antiderivative size = 1046, normalized size of antiderivative = 3.40 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \]
((a*(1 + Sin[e + f*x]))^(5/2)*(((A*d*(3*c^2 + 10*c*d + 19*d^2) - B*(15*c^3 + 30*c^2*d + 7*c*d^2 - 20*d^3))*((c + d)*(e + f*x - 2*Log[Sec[(e + f*x)/4 ]^2]) + Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x )/4]] - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4 ]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c + d]*Log[- #1 + Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 ^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c + d)^(7/2) + ((A*d*(3*c^2 + 1 0*c*d + 19*d^2) - B*(15*c^3 + 30*c^2*d + 7*c*d^2 - 20*d^3))*(-((c + d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2])) + Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c *#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] + d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x )/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + T an[(e + f*x)/4]]*#1 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqr t[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4] ]*#1^2 - 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + c*Sqrt[c + d]* Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c + d)^(7/2) - (4*Sqrt[d]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(15*B*c^...
Time = 1.52 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3454, 27, 3042, 3454, 27, 3042, 3460, 3042, 3252, 221}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int -\frac {(\sin (e+f x) a+a)^{3/2} (a (3 B c-7 A d-4 B d)-a (5 B c-A d+4 B d) \sin (e+f x))}{2 (c+d \sin (e+f x))^2}dx}{2 d (c+d)}+\frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\int \frac {(\sin (e+f x) a+a)^{3/2} (a (3 B c-7 A d-4 B d)-a (5 B c-A d+4 B d) \sin (e+f x))}{(c+d \sin (e+f x))^2}dx}{4 d (c+d)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\int \frac {(\sin (e+f x) a+a)^{3/2} (a (3 B c-7 A d-4 B d)-a (5 B c-A d+4 B d) \sin (e+f x))}{(c+d \sin (e+f x))^2}dx}{4 d (c+d)}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {\int -\frac {\sqrt {\sin (e+f x) a+a} \left (a^2 \left (A d (c+19 d)-B \left (5 c^2+3 d c-20 d^2\right )\right )-a^2 \left (3 A d (c+3 d)-B \left (15 c^2+25 d c+4 d^2\right )\right ) \sin (e+f x)\right )}{2 (c+d \sin (e+f x))}dx}{d (c+d)}+\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}}{4 d (c+d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 \left (A d (c+19 d)-B \left (5 c^2+3 d c-20 d^2\right )\right )-a^2 \left (3 A d (c+3 d)-B \left (15 c^2+25 d c+4 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)}dx}{2 d (c+d)}}{4 d (c+d)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 \left (A d (c+19 d)-B \left (5 c^2+3 d c-20 d^2\right )\right )-a^2 \left (3 A d (c+3 d)-B \left (15 c^2+25 d c+4 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)}dx}{2 d (c+d)}}{4 d (c+d)}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {a^2 \left (A d \left (3 c^2+10 c d+19 d^2\right )-B \left (15 c^3+30 c^2 d+7 c d^2-20 d^3\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}+\frac {2 a^3 \left (3 A d (c+3 d)-B \left (15 c^2+25 c d+4 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{2 d (c+d)}}{4 d (c+d)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {a^2 \left (A d \left (3 c^2+10 c d+19 d^2\right )-B \left (15 c^3+30 c^2 d+7 c d^2-20 d^3\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}+\frac {2 a^3 \left (3 A d (c+3 d)-B \left (15 c^2+25 c d+4 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{2 d (c+d)}}{4 d (c+d)}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {2 a^3 \left (3 A d (c+3 d)-B \left (15 c^2+25 c d+4 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}-\frac {2 a^3 \left (A d \left (3 c^2+10 c d+19 d^2\right )-B \left (15 c^3+30 c^2 d+7 c d^2-20 d^3\right )\right ) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f}}{2 d (c+d)}}{4 d (c+d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {\frac {a^2 \left (A d (c+7 d)-B \left (5 c^2+7 c d-4 d^2\right )\right ) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {2 a^3 \left (3 A d (c+3 d)-B \left (15 c^2+25 c d+4 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}-\frac {2 a^{5/2} \left (A d \left (3 c^2+10 c d+19 d^2\right )-B \left (15 c^3+30 c^2 d+7 c d^2-20 d^3\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt {c+d}}}{2 d (c+d)}}{4 d (c+d)}\) |
(a*(B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - ((a^2*(A*d*(c + 7*d) - B*(5*c^2 + 7*c*d - 4*d^2))*C os[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(d*(c + d)*f*(c + d*Sin[e + f*x])) - ((-2*a^(5/2)*(A*d*(3*c^2 + 10*c*d + 19*d^2) - B*(15*c^3 + 30*c^2*d + 7*c* d^2 - 20*d^3))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(3/2)*Sqrt[c + d]*f) + (2*a^3*(3*A*d*(c + 3*d) - B *(15*c^2 + 25*c*d + 4*d^2))*Cos[e + f*x])/(d*f*Sqrt[a + a*Sin[e + f*x]]))/ (2*d*(c + d)))/(4*d*(c + d))
3.4.6.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(1586\) vs. \(2(280)=560\).
Time = 63.26 (sec) , antiderivative size = 1587, normalized size of antiderivative = 5.15
-1/4*a*(-7*B*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f* x+e)^2*a^2*c*d^4+6*A*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2) )*sin(f*x+e)*a^2*c^3*d^2+13*A*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)* a*d^4+15*B*(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^4+10*A*arctanh( (-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^3*d^2+19*A*arctanh((- a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^2*d^3+4*B*(-a*(sin(f*x+ e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^4+19*A*arctanh((-a*(sin(f*x+e)-1))^(1/2 )*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a^2*d^5+20*B*arctanh((-a*(sin(f*x+e)-1 ))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*a^2*d^5+5*A*(-a*(sin(f*x+e)-1)) ^(3/2)*(a*(c+d)*d)^(1/2)*c^2*d^2+20*A*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/ (a*(c+d)*d)^(1/2))*sin(f*x+e)*a^2*c^2*d^3+38*A*arctanh((-a*(sin(f*x+e)-1)) ^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*a^2*c*d^4+8*B*(-a*(sin(f*x+e)-1))^( 1/2)*(a*(c+d)*d)^(1/2)*sin(f*x+e)^2*a*d^4-30*B*arctanh((-a*(sin(f*x+e)-1)) ^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*a^2*c^4*d+6*A*(-a*(sin(f*x+e)-1))^( 3/2)*(a*(c+d)*d)^(1/2)*c*d^3-15*a^2*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a *(c+d)*d)^(1/2))*B*c^5-4*B*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*d^4 -11*A*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*d^4-9*B*(-a*(sin(f*x+e)- 1))^(3/2)*(a*(c+d)*d)^(1/2)*c^3*d-2*B*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d )^(1/2)*c^2*d^2+15*B*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*c*d^3-30* B*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*c^4*d-7*B*...
Leaf count of result is larger than twice the leaf count of optimal. 1365 vs. \(2 (280) = 560\).
Time = 1.73 (sec) , antiderivative size = 3046, normalized size of antiderivative = 9.89 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]
[1/16*((15*B*a^2*c^5 - 3*(A - 20*B)*a^2*c^4*d - 2*(8*A - 41*B)*a^2*c^3*d^2 - 6*(7*A - 4*B)*a^2*c^2*d^3 - 3*(16*A + 11*B)*a^2*c*d^4 - (19*A + 20*B)*a ^2*d^5 - (15*B*a^2*c^3*d^2 - 3*(A - 10*B)*a^2*c^2*d^3 - (10*A - 7*B)*a^2*c *d^4 - (19*A + 20*B)*a^2*d^5)*cos(f*x + e)^3 - (30*B*a^2*c^4*d - 3*(2*A - 25*B)*a^2*c^3*d^2 - (23*A - 44*B)*a^2*c^2*d^3 - 3*(16*A + 11*B)*a^2*c*d^4 - (19*A + 20*B)*a^2*d^5)*cos(f*x + e)^2 + (15*B*a^2*c^5 - 3*(A - 10*B)*a^2 *c^4*d - 2*(5*A - 11*B)*a^2*c^3*d^2 - 2*(11*A - 5*B)*a^2*c^2*d^3 - (10*A - 7*B)*a^2*c*d^4 - (19*A + 20*B)*a^2*d^5)*cos(f*x + e) + (15*B*a^2*c^5 - 3* (A - 20*B)*a^2*c^4*d - 2*(8*A - 41*B)*a^2*c^3*d^2 - 6*(7*A - 4*B)*a^2*c^2* d^3 - 3*(16*A + 11*B)*a^2*c*d^4 - (19*A + 20*B)*a^2*d^5 - (15*B*a^2*c^3*d^ 2 - 3*(A - 10*B)*a^2*c^2*d^3 - (10*A - 7*B)*a^2*c*d^4 - (19*A + 20*B)*a^2* d^5)*cos(f*x + e)^2 + 2*(15*B*a^2*c^4*d - 3*(A - 10*B)*a^2*c^3*d^2 - (10*A - 7*B)*a^2*c^2*d^3 - (19*A + 20*B)*a^2*c*d^4)*cos(f*x + e))*sin(f*x + e)) *sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)* cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x...
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 895 vs. \(2 (280) = 560\).
Time = 0.41 (sec) , antiderivative size = 895, normalized size of antiderivative = 2.91 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]
1/8*sqrt(2)*(16*B*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/ 2*f*x + 1/2*e)/d^3 + sqrt(2)*(15*B*a^2*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 *e)) - 3*A*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 30*B*a^2*c^2*d* sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 10*A*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2 *f*x + 1/2*e)) + 7*B*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 19*A* a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 20*B*a^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt (-c*d - d^2))/((c^2*d^3 + 2*c*d^4 + d^5)*sqrt(-c*d - d^2)) - 2*(18*B*a^2*c ^3*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 10*A*a^2*c^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f *x + 1/2*e)^3 + 4*B*a^2*c^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1 /4*pi + 1/2*f*x + 1/2*e)^3 - 12*A*a^2*c*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 30*B*a^2*c*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 22*A*a^2*d^4*sgn(cos( -1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 8*B*a^2*d^4 *sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 7* B*a^2*c^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2* e) + 3*A*a^2*c^3*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f *x + 1/2*e) - 13*B*a^2*c^3*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4* pi + 1/2*f*x + 1/2*e) + 13*A*a^2*c^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/...
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]