3.325 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx\)
Optimal. Leaf size=261 \[ -\frac{\left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f (c-d)^3}-\frac{2 d^{3/2} (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{a^{5/2} f (c-d)^3 \sqrt{c+d}}-\frac{(3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac{(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]
[Out]
-((B*(5*c^2 - 34*c*d - 3*d^2) + A*(3*c^2 - 14*c*d + 43*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a +
a*Sin[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*(c - d)^3*f) - (2*d^(3/2)*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e +
f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(a^(5/2)*(c - d)^3*Sqrt[c + d]*f) - ((A - B)*Cos[e + f*x])/(4*(
c - d)*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*A*c + 5*B*c - 11*A*d + 3*B*d)*Cos[e + f*x])/(16*a*(c - d)^2*f*(a +
a*Sin[e + f*x])^(3/2))
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Rubi [A] time = 0.984072, antiderivative size = 261, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used =
{2978, 2985, 2649, 206, 2773, 208} \[ -\frac{\left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f (c-d)^3}-\frac{2 d^{3/2} (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{a^{5/2} f (c-d)^3 \sqrt{c+d}}-\frac{(3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac{(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])),x]
[Out]
-((B*(5*c^2 - 34*c*d - 3*d^2) + A*(3*c^2 - 14*c*d + 43*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a +
a*Sin[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*(c - d)^3*f) - (2*d^(3/2)*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e +
f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(a^(5/2)*(c - d)^3*Sqrt[c + d]*f) - ((A - B)*Cos[e + f*x])/(4*(
c - d)*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*A*c + 5*B*c - 11*A*d + 3*B*d)*Cos[e + f*x])/(16*a*(c - d)^2*f*(a +
a*Sin[e + f*x])^(3/2))
Rule 2978
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2985
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]
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])
Rule 2773
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx &=-\frac{(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac{\int \frac{-\frac{1}{2} a (3 A c+5 B c-8 A d)-\frac{3}{2} a (A-B) d \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx}{4 a^2 (c-d)}\\ &=-\frac{(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 \left (B c (5 c-29 d)+A \left (3 c^2-11 c d+32 d^2\right )\right )+\frac{1}{4} a^2 d (3 A c+5 B c-11 A d+3 B d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{8 a^4 (c-d)^2}\\ &=-\frac{(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac{\left (d^2 (B c-A d)\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3}+\frac{\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{32 a^2 (c-d)^3}\\ &=-\frac{(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}-\frac{\left (2 d^2 (B c-A d)\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d-d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{a^2 (c-d)^3 f}-\frac{\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 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 (c-d)^3 f}\\ &=-\frac{\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 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} (c-d)^3 f}-\frac{2 d^{3/2} (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{a^{5/2} (c-d)^3 \sqrt{c+d} f}-\frac{(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac{(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 5.34184, size = 550, normalized size = 2.11 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((1+i) (-1)^{3/4} \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+\frac{8 d^{3/2} (A d-B c) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (\sqrt{c+d}-\sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt{c+d}}+\frac{8 d^{3/2} (B c-A d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (\sqrt{c+d}+\sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )-\sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt{c+d}}+8 (A-B) (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-(c-d) (3 A c-11 A d+5 B c+3 B d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (c-d) (3 A c-11 A d+5 B c+3 B d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (B-A) (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{16 f (c-d)^3 (a (\sin (e+f x)+1))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(A - B)*(c - d)^2*Sin[(e + f*x)/2] + 4*(-A + B)*(c - d)^2*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]) + 2*(c - d)*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^2 - (c - d)*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 +
I)*(-1)^(3/4)*(B*(5*c^2 - 34*c*d - 3*d^2) + A*(3*c^2 - 14*c*d + 43*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 + (8*d^(3/2)*(-(B*c) + A*d)*(e + f*x - 2*Log[Sec[(e
+ f*x)/4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] + Sqrt[d]*Cos[(e + f*x)/2] - Sqrt[d]*Sin[(e + f*x)/2])])
*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)/Sqrt[c + d] + (8*d^(3/2)*(B*c - A*d)*(e + f*x - 2*Log[Sec[(e + f*x)/
4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] - Sqrt[d]*Cos[(e + f*x)/2] + Sqrt[d]*Sin[(e + f*x)/2])])*(Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2])^4)/Sqrt[c + d]))/(16*(c - d)^3*f*(a*(1 + Sin[e + f*x]))^(5/2))
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Maple [B] time = 2.361, size = 1418, normalized size = 5.4 \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))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x)
[Out]
1/32*(2*sin(f*x+e)*(64*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^3-64*B*a^(5/2)*arctan
h((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c*d^2-3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a
^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c^2+14*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)
^(1/2)*a^2*c*d-43*A*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*2^(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*(c+d)*d)^(1/2)*a^2*c^2+34*B*2^(1/2)*arctanh(1/2
*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c*d+3*B*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1
/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*2^(1/2)*a^2*d^2)+(-64*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a
^(5/2)*d^3+64*B*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c*d^2+3*A*2^(1/2)*arctanh(1/2*(a
-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c^2-14*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2
)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c*d+43*A*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)
*d)^(1/2)*2^(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*(c+d)*d)^(1/2)*a^
2*c^2-34*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c*d-3*B*arctanh(1
/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*2^(1/2)*a^2*d^2)*cos(f*x+e)^2+128*A*arctanh((a-a*
sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^3-128*B*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d
^2)^(1/2))*c*d^2-6*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c^2+28*
A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c*d-86*A*arctanh(1/2*(a-a*
sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*2^(1/2)*a^2*d^2-20*A*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(
1/2)*a^(3/2)*c^2+72*A*a^(3/2)*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c*d-52*A*a^(3/2)*(a-a*sin(f*x+e))^(1/2)
*(a*(c+d)*d)^(1/2)*d^2+6*A*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2-28*A*(a*(c+d)*d)^(1/2)*(a-a*si
n(f*x+e))^(3/2)*a^(1/2)*c*d+22*A*(a*(c+d)*d)^(1/2)*(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*(c+d)*d)^(1/2)*a^2*c^2+68*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(
1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a^2*c*d+6*B*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+
d)*d)^(1/2)*2^(1/2)*a^2*d^2-12*B*a^(3/2)*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*c^2-8*B*a^(3/2)*(a-a*sin(f*x
+e))^(1/2)*(a*(c+d)*d)^(1/2)*c*d+20*B*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^2+10*B*(a*(c+d)*d)^(1
/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2-4*B*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d-6*B*(a*(c+d)*d
)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^2)*(-a*(-1+sin(f*x+e)))^(1/2)/a^(9/2)/(a*(c+d)*d)^(1/2)/(1+sin(f*x+e)
)/(c-d)^3/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
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Maxima [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))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] time = 52.8022, size = 5966, normalized size = 22.86 \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))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="fricas")
[Out]
[1/64*(sqrt(2)*(((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)^3 - 4*(3*A + 5*B)*c^2 +
8*(7*A + 17*B)*c*d - 4*(43*A - 3*B)*d^2 + 3*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x
+ e)^2 - 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e) - (4*(3*A + 5*B)*c^2 - 8*(7
*A + 17*B)*c*d + 4*(43*A - 3*B)*d^2 - ((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)^2
+ 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*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)) - 32*(4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)*cos(f*x + e)^3 - 3*(B*a*c*d - A*a*d^2)*cos(f*x + e)^2
+ 2*(B*a*c*d - A*a*d^2)*cos(f*x + e) + (4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)*cos(f*x + e)^2 + 2*(B*a*c*
d - A*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f*x
+ e)^2 - c^2 - 2*c*d - d^2 - 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(
f*x + e) + (c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d/(a*c
+ a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^2)*co
s(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)
*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(4*(A - B)*c^
2 - 8*(A - B)*c*d + 4*(A - B)*d^2 + ((3*A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e)^2 + ((
7*A + B)*c^2 - 2*(11*A - 3*B)*c*d + (15*A - 7*B)*d^2)*cos(f*x + e) - (4*(A - B)*c^2 - 8*(A - B)*c*d + 4*(A - B
)*d^2 - ((3*A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e)
+ a))/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*
d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^
3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2
- 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 -
a^3*d^3)*f)*sin(f*x + e)), 1/64*(sqrt(2)*(((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x +
e)^3 - 4*(3*A + 5*B)*c^2 + 8*(7*A + 17*B)*c*d - 4*(43*A - 3*B)*d^2 + 3*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d
+ (43*A - 3*B)*d^2)*cos(f*x + e)^2 - 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)
- (4*(3*A + 5*B)*c^2 - 8*(7*A + 17*B)*c*d + 4*(43*A - 3*B)*d^2 - ((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A
- 3*B)*d^2)*cos(f*x + e)^2 + 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*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)*s
in(f*x + e) - cos(f*x + e) - 2)) + 64*(4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)*cos(f*x + e)^3 - 3*(B*a*c*d
- A*a*d^2)*cos(f*x + e)^2 + 2*(B*a*c*d - A*a*d^2)*cos(f*x + e) + (4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)
*cos(f*x + e)^2 + 2*(B*a*c*d - A*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin
(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) + 4*(4*(A - B)*c^2 - 8*(A - B
)*c*d + 4*(A - B)*d^2 + ((3*A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e)^2 + ((7*A + B)*c^2
- 2*(11*A - 3*B)*c*d + (15*A - 7*B)*d^2)*cos(f*x + e) - (4*(A - B)*c^2 - 8*(A - B)*c*d + 4*(A - B)*d^2 - ((3*
A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^3
*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^
3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^
2*d + 3*a^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^
3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f)
*sin(f*x + e))]
________________________________________________________________________________________
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))/(a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [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))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="giac")
[Out]
Timed out