Integrand size = 37, antiderivative size = 218 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\frac {2 a^{5/2} (c-d)^2 (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{7/2} \sqrt {c+d} f}+\frac {2 a^3 \left (5 A (3 c-7 d) d-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{15 d^3 f \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 (5 B c-5 A d-8 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 d^2 f}-\frac {2 a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 d f} \] Output:
2*a^(5/2)*(c-d)^2*(-A*d+B*c)*arctanh(a^(1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1/2 )/(a+a*sin(f*x+e))^(1/2))/d^(7/2)/(c+d)^(1/2)/f+2/15*a^3*(5*A*(3*c-7*d)*d- B*(15*c^2-35*c*d+32*d^2))*cos(f*x+e)/d^3/f/(a+a*sin(f*x+e))^(1/2)+2/15*a^2 *(-5*A*d+5*B*c-8*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d^2/f-2/5*a*B*cos( f*x+e)*(a+a*sin(f*x+e))^(3/2)/d/f
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 9.93 (sec) , antiderivative size = 992, normalized size of antiderivative = 4.55 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x]),x]
Output:
((a*(1 + Sin[e + f*x]))^(5/2)*(-30*Sqrt[d]*(A*d*(-2*c + 5*d) + B*(2*c^2 - 5*c*d + 5*d^2))*Cos[(e + f*x)/2] - 5*d^(3/2)*(-2*B*c + 2*A*d + 5*B*d)*Cos[ (3*(e + f*x))/2] + 3*B*d^(5/2)*Cos[(5*(e + f*x))/2] - (15*(c - d)^2*(B*c - A*d)*((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)^(3/2) + (15*(c - d)^2*(B*c - A*d)*((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]*L og[-#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 + ...
Time = 1.38 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3455, 27, 3042, 3455, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {2 \int \frac {(\sin (e+f x) a+a)^{3/2} (a (3 B c+5 A d)-a (5 B c-5 A d-8 B d) \sin (e+f x))}{2 (c+d \sin (e+f x))}dx}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{3/2} (a (3 B c+5 A d)-a (5 B c-5 A d-8 B d) \sin (e+f x))}{c+d \sin (e+f x)}dx}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{3/2} (a (3 B c+5 A d)-a (5 B c-5 A d-8 B d) \sin (e+f x))}{c+d \sin (e+f x)}dx}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2 \int -\frac {\sqrt {\sin (e+f x) a+a} \left ((B c (5 c-17 d)-5 A d (c+3 d)) a^2+\left (5 A (3 c-7 d) d-B \left (15 c^2-35 d c+32 d^2\right )\right ) \sin (e+f x) a^2\right )}{2 (c+d \sin (e+f x))}dx}{3 d}+\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}-\frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left ((B c (5 c-17 d)-5 A d (c+3 d)) a^2+\left (5 A (3 c-7 d) d-B \left (15 c^2-35 d c+32 d^2\right )\right ) \sin (e+f x) a^2\right )}{c+d \sin (e+f x)}dx}{3 d}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}-\frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left ((B c (5 c-17 d)-5 A d (c+3 d)) a^2+\left (5 A (3 c-7 d) d-B \left (15 c^2-35 d c+32 d^2\right )\right ) \sin (e+f x) a^2\right )}{c+d \sin (e+f x)}dx}{3 d}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}-\frac {\frac {15 a^2 (c-d)^2 (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}-\frac {2 a^3 \left (5 A d (3 c-7 d)-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}-\frac {\frac {15 a^2 (c-d)^2 (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}-\frac {2 a^3 \left (5 A d (3 c-7 d)-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}-\frac {-\frac {30 a^3 (c-d)^2 (B c-A d) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f}-\frac {2 a^3 \left (5 A d (3 c-7 d)-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a^2 (-5 A d+5 B c-8 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}-\frac {-\frac {30 a^{5/2} (c-d)^2 (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt {c+d}}-\frac {2 a^3 \left (5 A d (3 c-7 d)-B \left (15 c^2-35 c d+32 d^2\right )\right ) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}}{5 d}-\frac {2 a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 d f}\) |
Input:
Int[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x]) ,x]
Output:
(-2*a*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*d*f) + ((2*a^2*(5*B*c - 5*A*d - 8*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*d*f) - ((-30*a^ (5/2)*(c - d)^2*(B*c - A*d)*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*(5*A*(3 *c - 7*d)*d - B*(15*c^2 - 35*c*d + 32*d^2))*Cos[e + f*x])/(d*f*Sqrt[a + a* Sin[e + f*x]]))/(3*d))/(5*d)
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], 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] && 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. \(542\) vs. \(2(192)=384\).
Time = 2.60 (sec) , antiderivative size = 543, normalized size of antiderivative = 2.49
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-3 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {\left (c +d \right ) a d}\, d^{2}+5 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, a \,d^{2}-15 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{3} c^{2} d +30 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{3} c \,d^{2}-15 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{3} d^{3}-5 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, a c d +20 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, a \,d^{2}+15 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{3} c^{3}-30 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{3} c^{2} d +15 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{3} c \,d^{2}+15 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a^{2} c d -45 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a^{2} d^{2}-15 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a^{2} c^{2}+45 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a^{2} c d -60 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a^{2} d^{2}\right )}{15 d^{3} \sqrt {\left (c +d \right ) a d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(543\) |
Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x,method=_RET URNVERBOSE)
Output:
2/15*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(-3*B*(a-a*sin(f*x+e))^(5/2) *((c+d)*a*d)^(1/2)*d^2+5*A*(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*a*d^2- 15*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c^2*d+30*A* arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c*d^2-15*A*arcta nh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*d^3-5*B*(a-a*sin(f*x+ e))^(3/2)*((c+d)*a*d)^(1/2)*a*c*d+20*B*(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^ (1/2)*a*d^2+15*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3 *c^3-30*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c^2*d+ 15*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^3*c*d^2+15*A* (a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a^2*c*d-45*A*(a-a*sin(f*x+e))^(1/ 2)*((c+d)*a*d)^(1/2)*a^2*d^2-15*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2) *a^2*c^2+45*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a^2*c*d-60*B*(a-a*s in(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a^2*d^2)/d^3/((c+d)*a*d)^(1/2)/cos(f*x+ e)/(a+a*sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (192) = 384\).
Time = 1.14 (sec) , antiderivative size = 1314, normalized size of antiderivative = 6.03 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algo rithm="fricas")
Output:
[-1/30*(15*(B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A*a^2* d^3 + (B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A*a^2*d^3)* cos(f*x + e) + (B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A + B)*a^2*c*d^2 - A* a^2*d^3)*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)*co s(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 + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) - 4*(3*B*a^2*d^2*cos(f*x + e)^3 - 15*B*a^2*c^2 + 5*(3*A + 7*B)*a^2*c* d - (35*A + 32*B)*a^2*d^2 + (5*B*a^2*c*d - (5*A + 11*B)*a^2*d^2)*cos(f*x + e)^2 - (15*B*a^2*c^2 - 5*(3*A + 8*B)*a^2*c*d + 2*(20*A + 23*B)*a^2*d^2)*c os(f*x + e) - (3*B*a^2*d^2*cos(f*x + e)^2 - 15*B*a^2*c^2 + 5*(3*A + 7*B)*a ^2*c*d - (35*A + 32*B)*a^2*d^2 - (5*B*a^2*c*d - (5*A + 14*B)*a^2*d^2)*cos( f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^3*f*cos(f*x + e) + d^ 3*f*sin(f*x + e) + d^3*f), 1/15*(15*(B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2* A + B)*a^2*c*d^2 - A*a^2*d^3 + (B*a^2*c^3 - (A + 2*B)*a^2*c^2*d + (2*A ...
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{d \sin \left (f x + e\right ) + c} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algo rithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c), x)
Leaf count of result is larger than twice the leaf count of optimal. 523 vs. \(2 (192) = 384\).
Time = 0.27 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.40 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algo rithm="giac")
Output:
1/15*sqrt(2)*sqrt(a)*(15*sqrt(2)*(B*a^2*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 2*e)) - A*a^2*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*B*a^2*c^2*d*sg n(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*A*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f* x + 1/2*e)) + B*a^2*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - A*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))/(sqrt(-c*d - d^2)*d^3) + 2*(12*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)^5 + 10*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 - 10*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 - 40*B*a^2*d^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*p i + 1/2*f*x + 1/2*e)^3 + 15*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) - 15*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) - 45*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) + 45*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) + 60*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))/d^ 5)/f
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{c+d\,\sin \left (e+f\,x\right )} \,d x \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c + d*sin(e + f*x)) ,x)
Output:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c + d*sin(e + f*x)) , x)
\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\sqrt {a}\, a^{2} \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right ) d +c}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right ) d +c}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right ) d +c}d x \right ) a +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right ) d +c}d x \right ) b +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right ) d +c}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right ) d +c}d x \right ) b \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)*d + c),x)*a + int(( sqrt(sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x)*d + c),x)*b + int((s qrt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)*d + c),x)*a + 2*int(( sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)*d + c),x)*b + 2*int( (sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)*d + c),x)*a + int((sqr t(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)*d + c),x)*b)