\(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [93]

Optimal result

Integrand size = 38, antiderivative size = 161 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {4 \sqrt {2} a^2 (A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[Out]

-2/5*a^2*B*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)-2/3*a^2*(A+B)*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+4*a
^2*(A+B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^(1/2)/f/c^(1/2)-4*a^2*(A+B)*cos(f*x+
e)/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3046, 2939, 2758, 2728, 212} \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {4 \sqrt {2} a^2 (A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 c (A+B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}} \]

[In]

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

[Out]

(4*Sqrt[2]*a^2*(A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^
2*B*c^2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(5/2)) - (2*a^2*(A + B)*c*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e
+ f*x])^(3/2)) - (4*a^2*(A + B)*Cos[e + f*x])/(f*Sqrt[c - c*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 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2939

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx \\ & = -\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}+\left (a^2 (A+B) c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx \\ & = -\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\left (2 a^2 (A+B) c\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx \\ & = -\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\left (4 a^2 (A+B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = -\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {\left (8 a^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f} \\ & = \frac {4 \sqrt {2} a^2 (A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {\$Aborted} \]

[In]

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

[Out]

$Aborted

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.22

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (30 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) A +30 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) B -3 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}-5 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c -5 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c -30 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, A \,c^{2}-30 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, B \,c^{2}\right )}{15 c^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(197\)
parts	\(-\frac {A \,a^{2} \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}\, \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {B \,a^{2} \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (15 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-6 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}+10 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c -30 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{2}\right )}{15 c^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a^{2} \left (A +2 B \right ) \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-2 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a^{2} \left (2 A +B \right ) \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(409\)

[In]

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

[Out]

-2/15*(sin(f*x+e)-1)*(c*(1+sin(f*x+e)))^(1/2)*a^2*(30*c^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(
1/2)/c^(1/2))*A+30*c^(5/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*B-3*B*(c*(1+sin(f*x+e
)))^(5/2)-5*A*(c*(1+sin(f*x+e)))^(3/2)*c-5*B*(c*(1+sin(f*x+e)))^(3/2)*c-30*(c*(1+sin(f*x+e)))^(1/2)*A*c^2-30*(
c*(1+sin(f*x+e)))^(1/2)*B*c^2)/c^3/cos(f*x+e)/(c-c*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. 310 vs. \(2 (142) = 284\).

Time = 0.29 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.93 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, {\left (\frac {15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} c \cos \left (f x + e\right ) - {\left (A + B\right )} a^{2} c \sin \left (f x + e\right ) + {\left (A + B\right )} a^{2} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + {\left (3 \, B a^{2} \cos \left (f x + e\right )^{3} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - {\left (35 \, A + 41 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (10 \, A + 13 \, B\right )} a^{2} + {\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} - {\left (5 \, A + 11 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (10 \, A + 13 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{15 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \]

[In]

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

[Out]

2/15*(15*sqrt(2)*((A + B)*a^2*c*cos(f*x + e) - (A + B)*a^2*c*sin(f*x + e) + (A + B)*a^2*c)*log(-(cos(f*x + e)^
2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sq
rt(c) + 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(c) + (
3*B*a^2*cos(f*x + e)^3 + (5*A + 14*B)*a^2*cos(f*x + e)^2 - (35*A + 41*B)*a^2*cos(f*x + e) - 4*(10*A + 13*B)*a^
2 + (3*B*a^2*cos(f*x + e)^2 - (5*A + 11*B)*a^2*cos(f*x + e) - 4*(10*A + 13*B)*a^2)*sin(f*x + e))*sqrt(-c*sin(f
*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)

Sympy [F]

\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=a^{2} \left (\int \frac {A}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {2 A \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {A \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {2 B \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin ^{3}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(A/sqrt(-c*sin(e + f*x) + c), x) + Integral(2*A*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x) + Int
egral(A*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(B*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x)
+ Integral(2*B*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(B*sin(e + f*x)**3/sqrt(-c*sin(e + f*x)
+ c), x))

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (142) = 284\).

Time = 0.48 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.98 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, {\left (\frac {15 \, \sqrt {2} {\left (A a^{2} \sqrt {c} + B a^{2} \sqrt {c}\right )} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {8 \, \sqrt {2} {\left (10 \, A a^{2} \sqrt {c} + 13 \, B a^{2} \sqrt {c} - \frac {35 \, A a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {35 \, B a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {55 \, A a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {85 \, B a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {45 \, A a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} - \frac {45 \, B a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {15 \, A a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}} + \frac {30 \, B a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}\right )}}{c {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{5} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{15 \, f} \]

[In]

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

[Out]

2/15*(15*sqrt(2)*(A*a^2*sqrt(c) + B*a^2*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*
f*x + 1/2*e) + 1))/(c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 8*sqrt(2)*(10*A*a^2*sqrt(c) + 13*B*a^2*sqrt(c) -
35*A*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 35*B*a^2*sqrt(c)*
(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 55*A*a^2*sqrt(c)*(cos(-1/4*pi + 1/
2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 85*B*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e
) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 45*A*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos
(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 - 45*B*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/
2*f*x + 1/2*e) + 1)^3 + 15*A*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e
) + 1)^4 + 30*B*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4)/(c*
((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^5*sgn(sin(-1/4*pi + 1/2*f*x +
1/2*e))))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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