Integrand size = 37, antiderivative size = 200 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\sqrt {2} (A-B) (c-d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {4 \left (5 A (3 c-d) d+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 d (4 B c+5 A d-B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 a f}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a+a \sin (e+f x)}} \] Output:
-2^(1/2)*(A-B)*(c-d)^2*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x +e))^(1/2))/a^(1/2)/f-4/15*(5*A*(3*c-d)*d+B*(6*c^2-7*c*d+7*d^2))*cos(f*x+e )/f/(a+a*sin(f*x+e))^(1/2)-2/15*d*(5*A*d+4*B*c-B*d)*cos(f*x+e)*(a+a*sin(f* x+e))^(1/2)/a/f-2/5*B*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^(1/ 2)
Result contains complex when optimal does not.
Time = 3.63 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((60+60 i) (-1)^{3/4} (A-B) (c-d)^2 \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )-30 \left (A (4 c-d) d+2 B \left (c^2-c d+d^2\right )\right ) \cos \left (\frac {1}{2} (e+f x)\right )+5 d (-2 A d+B (-4 c+d)) \cos \left (\frac {3}{2} (e+f x)\right )+3 B d^2 \cos \left (\frac {5}{2} (e+f x)\right )+30 \left (A (4 c-d) d+2 B \left (c^2-c d+d^2\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )+5 d (-2 A d+B (-4 c+d)) \sin \left (\frac {3}{2} (e+f x)\right )-3 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{30 f \sqrt {a (1+\sin (e+f x))}} \] Input:
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/Sqrt[a + a*Sin[e + f*x]],x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((60 + 60*I)*(-1)^(3/4)*(A - B)*(c - d)^2*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])] - 30*(A*(4* c - d)*d + 2*B*(c^2 - c*d + d^2))*Cos[(e + f*x)/2] + 5*d*(-2*A*d + B*(-4*c + d))*Cos[(3*(e + f*x))/2] + 3*B*d^2*Cos[(5*(e + f*x))/2] + 30*(A*(4*c - d)*d + 2*B*(c^2 - c*d + d^2))*Sin[(e + f*x)/2] + 5*d*(-2*A*d + B*(-4*c + d ))*Sin[(3*(e + f*x))/2] - 3*B*d^2*Sin[(5*(e + f*x))/2]))/(30*f*Sqrt[a*(1 + Sin[e + f*x])])
Time = 1.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 3462, 27, 3042, 3447, 3042, 3502, 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.
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a \sin (e+f x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a \sin (e+f x)+a}}dx\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \frac {2 \int \frac {(c+d \sin (e+f x)) (a (5 A c-B c+4 B d)+a (4 B c+5 A d-B d) \sin (e+f x))}{2 \sqrt {\sin (e+f x) a+a}}dx}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (5 A c-B c+4 B d)+a (4 B c+5 A d-B d) \sin (e+f x))}{\sqrt {\sin (e+f x) a+a}}dx}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) (a (5 A c-B c+4 B d)+a (4 B c+5 A d-B d) \sin (e+f x))}{\sqrt {\sin (e+f x) a+a}}dx}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {a d (4 B c+5 A d-B d) \sin ^2(e+f x)+(a c (4 B c+5 A d-B d)+a d (5 A c-B c+4 B d)) \sin (e+f x)+a c (5 A c-B c+4 B d)}{\sqrt {\sin (e+f x) a+a}}dx}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a d (4 B c+5 A d-B d) \sin (e+f x)^2+(a c (4 B c+5 A d-B d)+a d (5 A c-B c+4 B d)) \sin (e+f x)+a c (5 A c-B c+4 B d)}{\sqrt {\sin (e+f x) a+a}}dx}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {2 \int \frac {\left (5 A \left (3 c^2+d^2\right )-B \left (3 c^2-16 d c+d^2\right )\right ) a^2+2 \left (5 A (3 c-d) d+B \left (6 c^2-7 d c+7 d^2\right )\right ) \sin (e+f x) a^2}{2 \sqrt {\sin (e+f x) a+a}}dx}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (5 A \left (3 c^2+d^2\right )-B \left (3 c^2-16 d c+d^2\right )\right ) a^2+2 \left (5 A (3 c-d) d+B \left (6 c^2-7 d c+7 d^2\right )\right ) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a}}dx}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (5 A \left (3 c^2+d^2\right )-B \left (3 c^2-16 d c+d^2\right )\right ) a^2+2 \left (5 A (3 c-d) d+B \left (6 c^2-7 d c+7 d^2\right )\right ) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a}}dx}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {15 a^2 (A-B) (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx-\frac {4 a^2 \left (5 A d (3 c-d)+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a^2 (A-B) (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx-\frac {4 a^2 \left (5 A d (3 c-d)+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {-\frac {30 a^2 (A-B) (c-d)^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}-\frac {4 a^2 \left (5 A d (3 c-d)+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {-\frac {15 \sqrt {2} a^{3/2} (A-B) (c-d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}-\frac {4 a^2 \left (5 A d (3 c-d)+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{3 a}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}}\) |
Input:
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/Sqrt[a + a*Sin[e + f*x]] ,x]
Output:
(-2*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*Sqrt[a + a*Sin[e + f*x]]) + ((-2*d*(4*B*c + 5*A*d - B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f ) + ((-15*Sqrt[2]*a^(3/2)*(A - B)*(c - d)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x]) /(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f - (4*a^2*(5*A*(3*c - d)*d + B*(6*c ^2 - 7*c*d + 7*d^2))*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]))/(3*a))/(5 *a)
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[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(395\) vs. \(2(179)=358\).
Time = 0.85 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.98
| method | result | size |
| default | \(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (15 A \sqrt {2}\, a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c^{2}-30 A \sqrt {2}\, a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c d +15 A \sqrt {2}\, a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d^{2}-15 B \sqrt {2}\, a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c^{2}+30 B \sqrt {2}\, a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c d -15 B \sqrt {2}\, a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d^{2}+6 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}} d^{2}-10 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a \,d^{2}-20 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a c d -10 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a \,d^{2}+60 A \,a^{2} c d \sqrt {a -a \sin \left (f x +e \right )}+30 a^{2} B \,c^{2} \sqrt {a -a \sin \left (f x +e \right )}+30 B \,a^{2} d^{2} \sqrt {a -a \sin \left (f x +e \right )}\right )}{15 a^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(396\) |
| parts | \(-\frac {A \,c^{2} \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {B \,d^{2} \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (15 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}}+10 a \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}}-30 a^{2} \sqrt {a -a \sin \left (f x +e \right )}\right )}{15 a^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {c \left (2 A d +B c \right ) \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-2 \sqrt {a -a \sin \left (f x +e \right )}\right )}{a \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}-\frac {d \left (A d +2 B c \right ) \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-2 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}}\right )}{3 a^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(419\) |
Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x,method=_R ETURNVERBOSE)
Output:
-1/15*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(15*A*2^(1/2)*a^(5/2)*arcta nh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c^2-30*A*2^(1/2)*a^(5/2)*ar ctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c*d+15*A*2^(1/2)*a^(5/2) *arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d^2-15*B*2^(1/2)*a^(5 /2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c^2+30*B*2^(1/2)*a ^(5/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c*d-15*B*2^(1/2 )*a^(5/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d^2+6*B*(a-a *sin(f*x+e))^(5/2)*d^2-10*A*(a-a*sin(f*x+e))^(3/2)*a*d^2-20*B*(a-a*sin(f*x +e))^(3/2)*a*c*d-10*B*(a-a*sin(f*x+e))^(3/2)*a*d^2+60*A*a^2*c*d*(a-a*sin(f *x+e))^(1/2)+30*a^2*B*c^2*(a-a*sin(f*x+e))^(1/2)+30*B*a^2*d^2*(a-a*sin(f*x +e))^(1/2))/a^3/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. 448 vs. \(2 (179) = 358\).
Time = 0.14 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.24 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left ({\left (A - B\right )} a c^{2} - 2 \, {\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2} + {\left ({\left (A - B\right )} a c^{2} - 2 \, {\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (A - B\right )} a c^{2} - 2 \, {\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2}\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {a}} - 4 \, {\left (3 \, B d^{2} \cos \left (f x + e\right )^{3} - 15 \, B c^{2} - 10 \, {\left (3 \, A - 2 \, B\right )} c d + {\left (10 \, A - 17 \, B\right )} d^{2} - {\left (10 \, B c d + {\left (5 \, A - 4 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, B c^{2} + 10 \, {\left (3 \, A - B\right )} c d - {\left (5 \, A - 16 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (3 \, B d^{2} \cos \left (f x + e\right )^{2} - 15 \, B c^{2} - 10 \, {\left (3 \, A - 2 \, B\right )} c d + {\left (10 \, A - 17 \, B\right )} d^{2} + {\left (10 \, B c d + {\left (5 \, A - B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{30 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, al gorithm="fricas")
Output:
-1/30*(15*sqrt(2)*((A - B)*a*c^2 - 2*(A - B)*a*c*d + (A - B)*a*d^2 + ((A - B)*a*c^2 - 2*(A - B)*a*c*d + (A - B)*a*d^2)*cos(f*x + e) + ((A - B)*a*c^2 - 2*(A - B)*a*c*d + (A - B)*a*d^2)*sin(f*x + e))*log(-(cos(f*x + e)^2 - ( cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f *x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) - 4*(3*B*d^ 2*cos(f*x + e)^3 - 15*B*c^2 - 10*(3*A - 2*B)*c*d + (10*A - 17*B)*d^2 - (10 *B*c*d + (5*A - 4*B)*d^2)*cos(f*x + e)^2 - (15*B*c^2 + 10*(3*A - B)*c*d - (5*A - 16*B)*d^2)*cos(f*x + e) - (3*B*d^2*cos(f*x + e)^2 - 15*B*c^2 - 10*( 3*A - 2*B)*c*d + (10*A - 17*B)*d^2 + (10*B*c*d + (5*A - B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e) + a*f*sin(f* x + e) + a*f)
\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {\left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(1/2),x)
Output:
Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))**2/sqrt(a*(sin(e + f*x) + 1)), x)
\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, al gorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2/sqrt(a*sin(f*x + e) + a), x)
Exception generated. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),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. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(1/ 2),x)
Output:
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(1/ 2), x)
\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )+1}d x \right ) a \,c^{2}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )+1}d x \right ) b \,d^{2}+\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )+1}d x \right ) a \,d^{2}+2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )+1}d x \right ) b c d +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right ) a c d +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right ) b \,c^{2}\right )}{a} \] Input:
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x)
Output:
(sqrt(a)*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x) + 1),x)*a*c**2 + int((s qrt(sin(e + f*x) + 1)*sin(e + f*x)**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) + 1),x)*a*d**2 + 2* int((sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(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) + 1),x)*a*c*d + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) + 1),x)*b*c**2))/ a