\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx\) [308]

Optimal result

Integrand size = 37, antiderivative size = 200 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\sqrt {2} (A-B) (c-d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {4 \left (5 A (3 c-d) d+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 d (4 B c+5 A d-B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 a f}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a+a \sin (e+f x)}} \]

[Out]

-(A-B)*(c-d)^2*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/f/a^(1/2)-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)-2/5*B*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*si
n(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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3062, 3047, 3102, 2830, 2728, 212} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\sqrt {2} (A-B) (c-d)^2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f}-\frac {4 \left (5 A d (3 c-d)+B \left (6 c^2-7 c d+7 d^2\right )\right ) \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {2 d (5 A d+4 B c-B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 a f}-\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a \sin (e+f x)+a}} \]

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

-((Sqrt[2]*(A - B)*(c - d)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f))
- (4*(5*A*(3*c - d)*d + B*(6*c^2 - 7*c*d + 7*d^2))*Cos[e + f*x])/(15*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]])/(15*a*f) - (2*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5
*f*Sqrt[a + a*Sin[e + f*x]])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

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] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[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)]

Rule 3047

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]

Rule 3062

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[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])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a+a \sin (e+f x)}}+\frac {2 \int \frac {(c+d \sin (e+f x)) \left (\frac {1}{2} a (5 A c-B c+4 B d)+\frac {1}{2} a (4 B c+5 A d-B d) \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{5 a} \\ & = -\frac {2 B \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt {a+a \sin (e+f x)}}+\frac {2 \int \frac {\frac {1}{2} a c (5 A c-B c+4 B d)+\left (\frac {1}{2} a c (4 B c+5 A d-B d)+\frac {1}{2} a d (5 A c-B c+4 B d)\right ) \sin (e+f x)+\frac {1}{2} a d (4 B c+5 A d-B d) \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{5 a} \\ & = -\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)}}+\frac {4 \int \frac {\frac {1}{4} a^2 \left (5 A \left (3 c^2+d^2\right )-B \left (3 c^2-16 c d+d^2\right )\right )+\frac {1}{2} a^2 \left (5 A (3 c-d) d+B \left (6 c^2-7 c d+7 d^2\right )\right ) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{15 a^2} \\ & = -\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)}}+\left ((A-B) (c-d)^2\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx \\ & = -\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)}}-\frac {\left (2 (A-B) (c-d)^2\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{f} \\ & = -\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)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.26 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((60+60 i) (-1)^{3/4} (A-B) (c-d)^2 \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )-30 \left (A (4 c-d) d+2 B \left (c^2-c d+d^2\right )\right ) \cos \left (\frac {1}{2} (e+f x)\right )+5 d (-2 A d+B (-4 c+d)) \cos \left (\frac {3}{2} (e+f x)\right )+3 B d^2 \cos \left (\frac {5}{2} (e+f x)\right )+30 \left (A (4 c-d) d+2 B \left (c^2-c d+d^2\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )+5 d (-2 A d+B (-4 c+d)) \sin \left (\frac {3}{2} (e+f x)\right )-3 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{30 f \sqrt {a (1+\sin (e+f x))}} \]

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

((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])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(395\) vs. \(2(179)=358\).

Time = 2.62 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.98

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (15 A \,a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c^{2}-30 A \,a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c d +15 A \,a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d^{2}-15 B \,a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c^{2}+30 B \,a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c d -15 B \,a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d^{2}+6 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}} d^{2}-10 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a \,d^{2}-20 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a c d -10 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a \,d^{2}+60 \sqrt {a -a \sin \left (f x +e \right )}\, A \,a^{2} c d +30 \sqrt {a -a \sin \left (f x +e \right )}\, B \,a^{2} c^{2}+30 \sqrt {a -a \sin \left (f x +e \right )}\, B \,a^{2} d^{2}\right )}{15 a^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(396\)
parts	\(-\frac {A \,c^{2} \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {c \left (2 d A +B c \right ) \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-2 \sqrt {a -a \sin \left (f x +e \right )}\right )}{a \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}-\frac {d \left (d A +2 B c \right ) \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-2 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}}\right )}{3 a^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {d^{2} B \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (15 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}}+10 a \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}}-30 a^{2} \sqrt {a -a \sin \left (f x +e \right )}\right )}{15 a^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(419\)

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/15*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(15*A*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2
)/a^(1/2))*c^2-30*A*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c*d+15*A*a^(5/2)*2^(1/
2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d^2-15*B*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(
1/2)*2^(1/2)/a^(1/2))*c^2+30*B*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c*d-15*B*a^
(5/2)*2^(1/2)*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*sin(f*x+e
))^(1/2)*A*a^2*c*d+30*(a-a*sin(f*x+e))^(1/2)*B*a^2*c^2+30*(a-a*sin(f*x+e))^(1/2)*B*a^2*d^2)/a^3/cos(f*x+e)/(a+
a*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (179) = 358\).

Time = 0.27 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.24 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left ({\left (A - B\right )} a c^{2} - 2 \, {\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2} + {\left ({\left (A - B\right )} a c^{2} - 2 \, {\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (A - B\right )} a c^{2} - 2 \, {\left (A - B\right )} a c d + {\left (A - B\right )} a d^{2}\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {a}} - 4 \, {\left (3 \, B d^{2} \cos \left (f x + e\right )^{3} - 15 \, B c^{2} - 10 \, {\left (3 \, A - 2 \, B\right )} c d + {\left (10 \, A - 17 \, B\right )} d^{2} - {\left (10 \, B c d + {\left (5 \, A - 4 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, B c^{2} + 10 \, {\left (3 \, A - B\right )} c d - {\left (5 \, A - 16 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (3 \, B d^{2} \cos \left (f x + e\right )^{2} - 15 \, B c^{2} - 10 \, {\left (3 \, A - 2 \, B\right )} c d + {\left (10 \, A - 17 \, B\right )} d^{2} + {\left (10 \, B c d + {\left (5 \, A - B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{30 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-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)

Sympy [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {\left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))**2/sqrt(a*(sin(e + f*x) + 1)), x)

Maxima [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2/sqrt(a*sin(f*x + e) + a), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (179) = 358\).

Time = 0.35 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (A \sqrt {a} c^{2} - B \sqrt {a} c^{2} - 2 \, A \sqrt {a} c d + 2 \, B \sqrt {a} c d + A \sqrt {a} d^{2} - B \sqrt {a} d^{2}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {15 \, \sqrt {2} {\left (A \sqrt {a} c^{2} - B \sqrt {a} c^{2} - 2 \, A \sqrt {a} c d + 2 \, B \sqrt {a} c d + A \sqrt {a} d^{2} - B \sqrt {a} d^{2}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {4 \, \sqrt {2} {\left (12 \, B a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 20 \, B a^{\frac {9}{2}} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 10 \, A a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 10 \, B a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B a^{\frac {9}{2}} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, A a^{\frac {9}{2}} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{30 \, f} \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/30*(15*sqrt(2)*(A*sqrt(a)*c^2 - B*sqrt(a)*c^2 - 2*A*sqrt(a)*c*d + 2*B*sqrt(a)*c*d + A*sqrt(a)*d^2 - B*sqrt(a
)*d^2)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 15*sqrt(2)*(A*sqrt(a)
*c^2 - B*sqrt(a)*c^2 - 2*A*sqrt(a)*c*d + 2*B*sqrt(a)*c*d + A*sqrt(a)*d^2 - B*sqrt(a)*d^2)*log(-sin(-1/4*pi + 1
/2*f*x + 1/2*e) + 1)/(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 4*sqrt(2)*(12*B*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f
*x + 1/2*e)^5 - 20*B*a^(9/2)*c*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 10*A*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f*x + 1
/2*e)^3 - 10*B*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 15*B*a^(9/2)*c^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)
+ 30*A*a^(9/2)*c*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 15*B*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^5*sgn(
cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(1/2),x)

[Out]

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^(1/2), x)