3.322 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=219 \[ -\frac{\left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{d^2 (A-9 B) \cos (e+f x)}{4 a^2 f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{(c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]
[Out]
-((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]*Sqrt[a +
a*Sin[e + f*x]])])/(16*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))
________________________________________________________________________________________
Rubi [A] time = 0.578671, antiderivative size = 219, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used =
{2977, 2968, 3019, 2751, 2649, 206} \[ -\frac{\left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{d^2 (A-9 B) \cos (e+f x)}{4 a^2 f \sqrt{a \sin (e+f x)+a}}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{(c-d) (3 A c+5 A d+5 B c-13 B d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[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]
-((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]*Sqrt[a +
a*Sin[e + f*x]])])/(16*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 2977
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 2968
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 3019
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]
Rule 2751
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*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)]
Rule 2649
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \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{(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 ) \operatorname{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 ) \tanh ^{-1}\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] time = 1.133, size = 544, normalized size = 2.48 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((2+2 i) (-1)^{3/4} \left (A \left (3 c^2+10 c d+19 d^2\right )+B \left (5 c^2+38 c d-75 d^2\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+11 A c^2 \sin \left (\frac{1}{2} (e+f x)\right )-3 A c^2 \sin \left (\frac{3}{2} (e+f x)\right )-11 A c^2 \cos \left (\frac{1}{2} (e+f x)\right )-3 A c^2 \cos \left (\frac{3}{2} (e+f x)\right )-6 A c d \sin \left (\frac{1}{2} (e+f x)\right )-10 A c d \sin \left (\frac{3}{2} (e+f x)\right )+6 A c d \cos \left (\frac{1}{2} (e+f x)\right )-10 A c d \cos \left (\frac{3}{2} (e+f x)\right )-5 A d^2 \sin \left (\frac{1}{2} (e+f x)\right )+13 A d^2 \sin \left (\frac{3}{2} (e+f x)\right )+5 A d^2 \cos \left (\frac{1}{2} (e+f x)\right )+13 A d^2 \cos \left (\frac{3}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac{1}{2} (e+f x)\right )-5 B c^2 \sin \left (\frac{3}{2} (e+f x)\right )+3 B c^2 \cos \left (\frac{1}{2} (e+f x)\right )-5 B c^2 \cos \left (\frac{3}{2} (e+f x)\right )-10 B c d \sin \left (\frac{1}{2} (e+f x)\right )+26 B c d \sin \left (\frac{3}{2} (e+f x)\right )+10 B c d \cos \left (\frac{1}{2} (e+f x)\right )+26 B c d \cos \left (\frac{3}{2} (e+f x)\right )+45 B d^2 \sin \left (\frac{1}{2} (e+f x)\right )-69 B d^2 \sin \left (\frac{3}{2} (e+f x)\right )-16 B d^2 \sin \left (\frac{5}{2} (e+f x)\right )-45 B d^2 \cos \left (\frac{1}{2} (e+f x)\right )-69 B d^2 \cos \left (\frac{3}{2} (e+f x)\right )+16 B d^2 \cos \left (\frac{5}{2} (e+f x)\right )\right )}{32 f (a (\sin (e+f x)+1))^{5/2}} \]
Antiderivative was successfully verified.
[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))
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Maple [B] time = 1.888, size = 982, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x)
[Out]
-1/32*(2*sin(f*x+e)*(3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+10*A*2^(1/2)*arct
anh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+19*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2
)/a^(1/2))*a^2*d^2+5*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+38*B*2^(1/2)*arctan
h(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-75*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/
a^(1/2))*a^2*d^2+64*B*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2))+(-3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(
1/2)/a^(1/2))*a^2*c^2-10*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-19*A*2^(1/2)*ar
ctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-5*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/
2)/a^(1/2))*a^2*c^2-38*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+75*B*2^(1/2)*arct
anh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-64*B*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2))*cos(f*x+e)^2+
6*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+20*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+
e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+38*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+20
*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2+24*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d-44*A*d^2*a^(3/2)*(a-a*sin(f*x+e)
)^(1/2)-6*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2-20*A*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d+26*A*(a-a*sin(f*x+e))^(
3/2)*a^(1/2)*d^2+10*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+76*B*2^(1/2)*arctanh
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-150*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/
a^(1/2))*a^2*d^2+12*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2-88*B*c*d*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+204*B*d^2*a^(
3/2)*(a-a*sin(f*x+e))^(1/2)-10*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2+52*B*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d-42
*B*d^2*(a-a*sin(f*x+e))^(3/2)*a^(1/2))*(-a*(-1+sin(f*x+e)))^(1/2)/a^(9/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f
*x+e))^(1/2)/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Fricas [B] time = 2.33657, size = 1817, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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]
sage2