Integrand size = 37, antiderivative size = 219 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) (3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(A-9 B) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}} \] Output:
-1/32*(B*(5*c^2+38*c*d-75*d^2)+A*(3*c^2+10*c*d+19*d^2))*arctanh(1/2*a^(1/2 )*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/f-1/16*(c-d)* (3*A*c+5*A*d+5*B*c-13*B*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)+1/4*(A-9* B)*d^2*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)-1/4*(A-B)*cos(f*x+e)*(c+d*s in(f*x+e))^2/f/(a+a*sin(f*x+e))^(5/2)
Result contains complex when optimal does not.
Time = 4.67 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.48 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-11 A c^2 \cos \left (\frac {1}{2} (e+f x)\right )+3 B c^2 \cos \left (\frac {1}{2} (e+f x)\right )+6 A c d \cos \left (\frac {1}{2} (e+f x)\right )+10 B c d \cos \left (\frac {1}{2} (e+f x)\right )+5 A d^2 \cos \left (\frac {1}{2} (e+f x)\right )-45 B d^2 \cos \left (\frac {1}{2} (e+f x)\right )-3 A c^2 \cos \left (\frac {3}{2} (e+f x)\right )-5 B c^2 \cos \left (\frac {3}{2} (e+f x)\right )-10 A c d \cos \left (\frac {3}{2} (e+f x)\right )+26 B c d \cos \left (\frac {3}{2} (e+f x)\right )+13 A d^2 \cos \left (\frac {3}{2} (e+f x)\right )-69 B d^2 \cos \left (\frac {3}{2} (e+f x)\right )+16 B d^2 \cos \left (\frac {5}{2} (e+f x)\right )+11 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )-6 A c d \sin \left (\frac {1}{2} (e+f x)\right )-10 B c d \sin \left (\frac {1}{2} (e+f x)\right )-5 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )+45 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )+(2+2 i) (-1)^{3/4} \left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 d^2\right )\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-3 A c^2 \sin \left (\frac {3}{2} (e+f x)\right )-5 B c^2 \sin \left (\frac {3}{2} (e+f x)\right )-10 A c d \sin \left (\frac {3}{2} (e+f x)\right )+26 B c d \sin \left (\frac {3}{2} (e+f x)\right )+13 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )-69 B d^2 \sin \left (\frac {3}{2} (e+f x)\right )-16 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{32 f (a (1+\sin (e+f x)))^{5/2}} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x ])^(5/2),x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-11*A*c^2*Cos[(e + f*x)/2] + 3*B*c ^2*Cos[(e + f*x)/2] + 6*A*c*d*Cos[(e + f*x)/2] + 10*B*c*d*Cos[(e + f*x)/2] + 5*A*d^2*Cos[(e + f*x)/2] - 45*B*d^2*Cos[(e + f*x)/2] - 3*A*c^2*Cos[(3*( e + f*x))/2] - 5*B*c^2*Cos[(3*(e + f*x))/2] - 10*A*c*d*Cos[(3*(e + f*x))/2 ] + 26*B*c*d*Cos[(3*(e + f*x))/2] + 13*A*d^2*Cos[(3*(e + f*x))/2] - 69*B*d ^2*Cos[(3*(e + f*x))/2] + 16*B*d^2*Cos[(5*(e + f*x))/2] + 11*A*c^2*Sin[(e + f*x)/2] - 3*B*c^2*Sin[(e + f*x)/2] - 6*A*c*d*Sin[(e + f*x)/2] - 10*B*c*d *Sin[(e + f*x)/2] - 5*A*d^2*Sin[(e + f*x)/2] + 45*B*d^2*Sin[(e + f*x)/2] + (2 + 2*I)*(-1)^(3/4)*(B*(5*c^2 + 38*c*d - 75*d^2) + A*(3*c^2 + 10*c*d + 1 9*d^2))*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 3*A*c^2*Sin[(3*(e + f*x))/2] - 5*B*c^2*Sin [(3*(e + f*x))/2] - 10*A*c*d*Sin[(3*(e + f*x))/2] + 26*B*c*d*Sin[(3*(e + f *x))/2] + 13*A*d^2*Sin[(3*(e + f*x))/2] - 69*B*d^2*Sin[(3*(e + f*x))/2] - 16*B*d^2*Sin[(5*(e + f*x))/2]))/(32*f*(a*(1 + Sin[e + f*x]))^(5/2))
Time = 1.25 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 3456, 27, 3042, 3447, 3042, 3498, 27, 3042, 3230, 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.
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (3 A c+5 B c+4 A d-4 B d)-a (A-9 B) d \sin (e+f x))}{2 (\sin (e+f x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (3 A c+5 B c+4 A d-4 B d)-a (A-9 B) d \sin (e+f x))}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (3 A c+5 B c+4 A d-4 B d)-a (A-9 B) d \sin (e+f x))}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {-a (A-9 B) d^2 \sin ^2(e+f x)+(a d (3 A c+5 B c+4 A d-4 B d)-a (A-9 B) c d) \sin (e+f x)+a c (3 A c+5 B c+4 A d-4 B d)}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-a (A-9 B) d^2 \sin (e+f x)^2+(a d (3 A c+5 B c+4 A d-4 B d)-a (A-9 B) c d) \sin (e+f x)+a c (3 A c+5 B c+4 A d-4 B d)}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {-\frac {\int -\frac {a^2 \left (B \left (5 c^2+38 d c-39 d^2\right )+A \left (3 c^2+10 d c+15 d^2\right )\right )-4 a^2 (A-9 B) d^2 \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 \left (B \left (5 c^2+38 d c-39 d^2\right )+A \left (3 c^2+10 d c+15 d^2\right )\right )-4 a^2 (A-9 B) d^2 \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 \left (B \left (5 c^2+38 d c-39 d^2\right )+A \left (3 c^2+10 d c+15 d^2\right )\right )-4 a^2 (A-9 B) d^2 \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {a^2 \left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {8 a^2 d^2 (A-9 B) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 \left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {8 a^2 d^2 (A-9 B) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {\frac {8 a^2 d^2 (A-9 B) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 \left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{4 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {8 a^2 d^2 (A-9 B) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {\sqrt {2} a^{3/2} \left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{4 a^2}-\frac {a (c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
Input:
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^(5/ 2),x]
Output:
-1/4*((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]) ^(5/2)) + (-1/2*(a*(c - d)*(3*A*c + 5*B*c + 5*A*d - 13*B*d)*Cos[e + f*x])/ (f*(a + a*Sin[e + f*x])^(3/2)) + (-((Sqrt[2]*a^(3/2)*(B*(5*c^2 + 38*c*d - 75*d^2) + A*(3*c^2 + 10*c*d + 19*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqr t[2]*Sqrt[a + a*Sin[e + f*x]])])/f) + (8*a^2*(A - 9*B)*d^2*Cos[e + f*x])/( f*Sqrt[a + a*Sin[e + f*x]]))/(4*a^2))/(8*a^2)
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[(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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{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] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(851\) vs. \(2(196)=392\).
Time = 1.00 (sec) , antiderivative size = 852, normalized size of antiderivative = 3.89
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(852\) |
| default | \(\text {Expression too large to display}\) | \(982\) |
Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x,method=_R ETURNVERBOSE)
Output:
-1/32*A*c^2/a^(9/2)*(-3*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2) /a^(1/2))*a^2*cos(f*x+e)^2+6*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^ (1/2)/a^(1/2))*a^2*sin(f*x+e)+6*a^(3/2)*(a-a*sin(f*x+e))^(1/2)*sin(f*x+e)+ 6*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2+14*(a-a* sin(f*x+e))^(1/2)*a^(3/2))*(-a*(sin(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/cos(f* x+e)/(a+a*sin(f*x+e))^(1/2)/f+1/32*B*d^2/a^(9/2)*(-75*2^(1/2)*arctanh(1/2* (a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*cos(f*x+e)^2+64*a^(3/2)*(a-a*s in(f*x+e))^(1/2)*cos(f*x+e)^2+150*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/ 2)*2^(1/2)/a^(1/2))*a^2*sin(f*x+e)-128*a^(3/2)*(a-a*sin(f*x+e))^(1/2)*sin( f*x+e)+150*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2 -204*(a-a*sin(f*x+e))^(1/2)*a^(3/2)+42*(a-a*sin(f*x+e))^(3/2)*a^(1/2))*(-a *(sin(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f- 1/32*c*(2*A*d+B*c)/a^(11/2)*(-5*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2) *2^(1/2)/a^(1/2))*a^3*cos(f*x+e)^2+10*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e)) ^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a^3+12*a^(5/2)*(a-a*sin(f*x+e))^(1/2)-1 0*(a-a*sin(f*x+e))^(3/2)*a^(3/2)+10*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^( 1/2)*2^(1/2)/a^(1/2))*a^3)*(-a*(sin(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/cos(f* x+e)/(a+a*sin(f*x+e))^(1/2)/f+1/32*d*(A*d+2*B*c)*(19*2^(1/2)*arctanh(1/2*( a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*cos(f*x+e)^2-38*2^(1/2)*arctanh (1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(f*x+e)+44*(a-a*sin...
Leaf count of result is larger than twice the leaf count of optimal. 744 vs. \(2 (196) = 392\).
Time = 0.10 (sec) , antiderivative size = 744, normalized size of antiderivative = 3.40 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, al gorithm="fricas")
Output:
-1/64*(sqrt(2)*(((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2) *cos(f*x + e)^3 - 4*(3*A + 5*B)*c^2 - 8*(5*A + 19*B)*c*d - 4*(19*A - 75*B) *d^2 + 3*((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(f* x + e)^2 - 2*((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*co s(f*x + e) - (4*(3*A + 5*B)*c^2 + 8*(5*A + 19*B)*c*d + 4*(19*A - 75*B)*d^2 - ((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(f*x + e) ^2 + 2*((3*A + 5*B)*c^2 + 2*(5*A + 19*B)*c*d + (19*A - 75*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(a*sin( f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(32*B*d^2*cos(f*x + e)^3 - 4*(A - B)*c^2 + 8*(A - B)*c*d - 4*(A - B)*d^2 - ((3*A + 5*B)*c^2 + 2*(5*A - 13*B)*c*d - (13*A - 53*B)*d^2)*cos(f*x + e)^2 - ((7*A + B)*c^2 + 2*(A - 9*B)*c*d - 9*(A - 9*B)*d^2)*cos(f*x + e) - (32*B*d^2*cos(f*x + e)^2 - 4*(A - B)*c^2 + 8*(A - B)*c*d - 4*(A - B)*d^2 + ((3*A + 5*B)*c^2 + 2*(5*A - 13 *B)*c*d - (13*A - 85*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*s in(f*x + e))
Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, al gorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2/(a*sin(f*x + e) + a) ^(5/2), x)
Exception generated. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, al gorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(5/ 2),x)
Output:
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(5/ 2), x)
\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a \,c^{2}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b \,d^{2}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a \,d^{2}+2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b c d +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) a c d +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right ) b \,c^{2}\right )}{a^{3}} \] Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x)
Output:
(sqrt(a)*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a*c**2 + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x) **3)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*b*d**2 + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a*d**2 + 2*int((sqrt(sin(e + f*x) + 1)* sin(e + f*x)**2)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1 ),x)*b*c*d + 2*int((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*a*c*d + int((sqrt(sin(e + f*x ) + 1)*sin(e + f*x))/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x)*b*c**2))/a**3