\(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) [95]

Optimal result

Integrand size = 38, antiderivative size = 175 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 a^2 (A+9 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/4*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(9/2)-1/8*a^2*(A+9*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(5/2)+
3/8*a^2*(A+9*B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)-3/8*a^2*(A+9*
B)*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3046, 2938, 2759, 2758, 2728, 212} \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 a^2 (A+9 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(3*a^2*(A + 9*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(4*Sqrt[2]*c^(5/2)*f) + (
a^2*(A + B)*c^2*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*(A + 9*B)*Cos[e + f*x]^3)/(8*f*(c - c*
Sin[e + f*x])^(5/2)) - (3*a^2*(A + 9*B)*Cos[e + f*x])/(8*c^2*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 2759

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

Rule 2938

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p
+ 1))), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*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))^{9/2}} \, dx \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {1}{8} \left (a^2 (A+9 B) c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}+\frac {\left (3 a^2 (A+9 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 c} \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (3 a^2 (A+9 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 c^2} \\ & = \frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (3 a^2 (A+9 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 )}{4 c^2 f} \\ & = \frac {3 a^2 (A+9 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.51 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.97 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (4 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-(5 A+13 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(3+3 i) \sqrt [4]{-1} (A+9 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-8 B \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+8 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-2 (5 A+13 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )-8 B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^{5/2}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - (5*A + 13*B)*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (3 + 3*I)*(-1)^(1/4)*(A + 9*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(
e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 8*B*Cos[(e + f*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)
/2])^4 + 8*(A + B)*Sin[(e + f*x)/2] - 2*(5*A + 13*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]
- 8*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(4*f*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(152)=304\).

Time = 3.30 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.21

method	result	size
default	\(-\frac {a^{2} \left (3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}+27 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-6 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}-16 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-54 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}+10 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}+3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+26 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}+32 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, B \,c^{\frac {3}{2}} \sin \left (f x +e \right )+27 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-12 A \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}-60 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{8 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(386\)
parts	\(\text {Expression too large to display}\)	\(828\)

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/8/c^(9/2)*a^2*(3*A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2+27*B*2^(1
/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-6*A*2^(1/2)*arctanh(1/2*(c*(1+sin(f
*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^2-16*B*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)*sin(f*x+e)^2-54*B*2^(1/2)*
arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^2+10*A*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+3*A
*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2+26*B*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+32*(c
*(1+sin(f*x+e)))^(1/2)*B*c^(3/2)*sin(f*x+e)+27*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))
*c^2-12*A*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)-60*B*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2))*(c*(1+sin(f*x+e)))^(1/2)/(si
n(f*x+e)-1)/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. 449 vs. \(2 (152) = 304\).

Time = 0.28 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.57 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (A + 9 \, B\right )} a^{2} - {\left ({\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (A + 9 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (8 \, B a^{2} \cos \left (f x + e\right )^{3} - {\left (5 \, A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - {\left (A + 25 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \, {\left (A + B\right )} a^{2} + {\left (8 \, B a^{2} \cos \left (f x + e\right )^{2} + {\left (5 \, A + 29 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/16*(3*sqrt(2)*((A + 9*B)*a^2*cos(f*x + e)^3 + 3*(A + 9*B)*a^2*cos(f*x + e)^2 - 2*(A + 9*B)*a^2*cos(f*x + e)
- 4*(A + 9*B)*a^2 - ((A + 9*B)*a^2*cos(f*x + e)^2 - 2*(A + 9*B)*a^2*cos(f*x + e) - 4*(A + 9*B)*a^2)*sin(f*x +
e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e)
+ 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(
f*x + e) - cos(f*x + e) - 2)) - 4*(8*B*a^2*cos(f*x + e)^3 - (5*A + 21*B)*a^2*cos(f*x + e)^2 - (A + 25*B)*a^2*c
os(f*x + e) + 4*(A + B)*a^2 + (8*B*a^2*cos(f*x + e)^2 + (5*A + 29*B)*a^2*cos(f*x + e) + 4*(A + B)*a^2)*sin(f*x
+ e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^
3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(5/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 601 vs. \(2 (152) = 304\).

Time = 0.67 (sec) , antiderivative size = 601, normalized size of antiderivative = 3.43 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-1/64*(256*sqrt(2)*B*a^2/(c^(5/2)*((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)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 12*sqrt(2)*(A*a^2*sqrt(c) + 9*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^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*
(A*a^2*sqrt(c) + B*a^2*sqrt(c) + 8*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) + 24*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) + 18
*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 + 162*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)*(cos(-1/4*pi + 1/2*f*x + 1/2
*e) + 1)^2/(c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - (8*sqrt(2)*A*a^2
*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2
*e) + 1) + 24*sqrt(2)*B*a^2*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(
cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + sqrt(2)*A*a^2*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/
4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + sqrt(2)*B*a^2*c^(7/2)*(cos(-1/4*pi + 1/2*f*x
+ 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/c^6)/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(5/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(5/2), x)