\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx\) [322]

Optimal result

Integrand size = 37, antiderivative size = 219 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) (3 A c+5 B c+5 A d-13 B d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(A-9 B) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 219, 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 = {3056, 3047, 3098, 2830, 2728, 212} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (A-9 B) \cos (e+f x)}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {(c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

[In]

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

[Out]

-1/16*((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])/(Sqrt[2]*Sqr
t[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(5/2)*f) - ((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)) + ((A - 9*B)*d^2*Cos[e + f*x])/(4*a^2*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[
e + f*x]*(c + d*Sin[e + f*x])^2)/(4*f*(a + a*Sin[e + f*x])^(5/2))

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 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] - Dist[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] /; FreeQ[{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] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3098

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] + D
ist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {(c+d \sin (e+f x)) \left (\frac {1}{2} a (3 A c+5 B c+4 A d-4 B d)-\frac {1}{2} a (A-9 B) d \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a c (3 A c+5 B c+4 A d-4 B d)+\left (-\frac {1}{2} a (A-9 B) c d+\frac {1}{2} a d (3 A c+5 B c+4 A d-4 B d)\right ) \sin (e+f x)-\frac {1}{2} a (A-9 B) d^2 \sin ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2} \\ & = -\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-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{4} a^2 \left (B \left (5 c^2+38 c d-39 d^2\right )+A \left (3 c^2+10 c d+15 d^2\right )\right )+a^2 (A-9 B) d^2 \sin (e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{8 a^4} \\ & = -\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}}+\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 ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2} \\ & = -\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}}-\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 {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 )}{16 a^2 f} \\ & = -\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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.86 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.48 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-11 A c^2 \cos \left (\frac {1}{2} (e+f x)\right )+3 B c^2 \cos \left (\frac {1}{2} (e+f x)\right )+6 A c d \cos \left (\frac {1}{2} (e+f x)\right )+10 B c d \cos \left (\frac {1}{2} (e+f x)\right )+5 A d^2 \cos \left (\frac {1}{2} (e+f x)\right )-45 B d^2 \cos \left (\frac {1}{2} (e+f x)\right )-3 A c^2 \cos \left (\frac {3}{2} (e+f x)\right )-5 B c^2 \cos \left (\frac {3}{2} (e+f x)\right )-10 A c d \cos \left (\frac {3}{2} (e+f x)\right )+26 B c d \cos \left (\frac {3}{2} (e+f x)\right )+13 A d^2 \cos \left (\frac {3}{2} (e+f x)\right )-69 B d^2 \cos \left (\frac {3}{2} (e+f x)\right )+16 B d^2 \cos \left (\frac {5}{2} (e+f x)\right )+11 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )-6 A c d \sin \left (\frac {1}{2} (e+f x)\right )-10 B c d \sin \left (\frac {1}{2} (e+f x)\right )-5 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )+45 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )+(2+2 i) (-1)^{3/4} \left (B \left (5 c^2+38 c d-75 d^2\right )+A \left (3 c^2+10 c d+19 d^2\right )\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-3 A c^2 \sin \left (\frac {3}{2} (e+f x)\right )-5 B c^2 \sin \left (\frac {3}{2} (e+f x)\right )-10 A c d \sin \left (\frac {3}{2} (e+f x)\right )+26 B c d \sin \left (\frac {3}{2} (e+f x)\right )+13 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )-69 B d^2 \sin \left (\frac {3}{2} (e+f x)\right )-16 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{32 f (a (1+\sin (e+f x)))^{5/2}} \]

[In]

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

[Out]

((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 + 19*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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(851\) vs. \(2(196)=392\).

Time = 3.22 (sec) , antiderivative size = 852, normalized size of antiderivative = 3.89

method	result	size
parts	\(\text {Expression too large to display}\)	\(852\)
default	\(\text {Expression too large to display}\)	\(982\)

[In]

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

[Out]

-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-a*sin(f*x+e))^(1/2)*sin(f*x+e)*a^(3/2
)+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*(si
n(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)*(-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-a*sin(f*x+e))^(1/2)*a^(5/2)-10*(a-a*sin(f*x+e))^(3/2)*a^(3/2)+10*2^(1/2)*arc
tanh(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)/a^(11/2)/(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)/a^(9/2)*(-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)+38*
2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2-44*(a-a*sin(f*x+e))^(1/2)*a^(3/2)+26*(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*d^2*B
/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*sin
(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-a*sin(f*x+e))^(1/2)*sin(f*x+e)*a^(3/2)+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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 744 vs. \(2 (196) = 392\).

Time = 0.29 (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=-\frac {\sqrt {2} {\left ({\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 4 \, {\left (3 \, A + 5 \, B\right )} c^{2} - 8 \, {\left (5 \, A + 19 \, B\right )} c d - 4 \, {\left (19 \, A - 75 \, B\right )} d^{2} + 3 \, {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, {\left (3 \, A + 5 \, B\right )} c^{2} + 8 \, {\left (5 \, A + 19 \, B\right )} c d + 4 \, {\left (19 \, A - 75 \, B\right )} d^{2} - {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A + 19 \, B\right )} c d + {\left (19 \, A - 75 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 (32 \, B d^{2} \cos \left (f x + e\right )^{3} - 4 \, {\left (A - B\right )} c^{2} + 8 \, {\left (A - B\right )} c d - 4 \, {\left (A - B\right )} d^{2} - {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A - 13 \, B\right )} c d - {\left (13 \, A - 53 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left ({\left (7 \, A + B\right )} c^{2} + 2 \, {\left (A - 9 \, B\right )} c d - 9 \, {\left (A - 9 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (32 \, B d^{2} \cos \left (f x + e\right )^{2} - 4 \, {\left (A - B\right )} c^{2} + 8 \, {\left (A - B\right )} c d - 4 \, {\left (A - B\right )} d^{2} + {\left ({\left (3 \, A + 5 \, B\right )} c^{2} + 2 \, {\left (5 \, A - 13 \, B\right )} c d - {\left (13 \, A - 85 \, 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}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

-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)*cos(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)*sin(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (196) = 392\).

Time = 0.43 (sec) , antiderivative size = 530, normalized size of antiderivative = 2.42 \[ \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 {\frac {128 \, \sqrt {2} B d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (3 \, A \sqrt {a} c^{2} + 5 \, B \sqrt {a} c^{2} + 10 \, A \sqrt {a} c d + 38 \, B \sqrt {a} c d + 19 \, A \sqrt {a} d^{2} - 75 \, 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^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (3 \, A \sqrt {a} c^{2} + 5 \, B \sqrt {a} c^{2} + 10 \, A \sqrt {a} c d + 38 \, B \sqrt {a} c d + 19 \, A \sqrt {a} d^{2} - 75 \, 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^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, \sqrt {2} {\left (3 \, A \sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, B \sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 10 \, A \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 26 \, B \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 13 \, A \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 21 \, B \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, A \sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B \sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, A \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 22 \, B \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 11 \, A \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 19 \, B \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{64 \, f} \]

[In]

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

[Out]

1/64*(128*sqrt(2)*B*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)
*(3*A*sqrt(a)*c^2 + 5*B*sqrt(a)*c^2 + 10*A*sqrt(a)*c*d + 38*B*sqrt(a)*c*d + 19*A*sqrt(a)*d^2 - 75*B*sqrt(a)*d^
2)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(3*A*sqrt(a)*c^
2 + 5*B*sqrt(a)*c^2 + 10*A*sqrt(a)*c*d + 38*B*sqrt(a)*c*d + 19*A*sqrt(a)*d^2 - 75*B*sqrt(a)*d^2)*log(-sin(-1/4
*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*sqrt(2)*(3*A*sqrt(a)*c^2*sin(-1/4*pi
+ 1/2*f*x + 1/2*e)^3 + 5*B*sqrt(a)*c^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 10*A*sqrt(a)*c*d*sin(-1/4*pi + 1/2*
f*x + 1/2*e)^3 - 26*B*sqrt(a)*c*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 13*A*sqrt(a)*d^2*sin(-1/4*pi + 1/2*f*x +
1/2*e)^3 + 21*B*sqrt(a)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*A*sqrt(a)*c^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)
- 3*B*sqrt(a)*c^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 6*A*sqrt(a)*c*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 22*B*sqrt(
a)*c*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 11*A*sqrt(a)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 19*B*sqrt(a)*d^2*sin
(-1/4*pi + 1/2*f*x + 1/2*e))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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