3.292 \(\int \frac{\sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=192 \[ -\frac{\sqrt{a} (3 A d+B (c+4 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 d^{3/2} f (c+d)^{5/2}}-\frac{a (3 A d+B (c+4 d)) \cos (e+f x)}{4 d f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac{a (B c-A d) \cos (e+f x)}{2 d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^2} \]
[Out]
-(Sqrt[a]*(3*A*d + B*(c + 4*d))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])]
)/(4*d^(3/2)*(c + d)^(5/2)*f) + (a*(B*c - A*d)*Cos[e + f*x])/(2*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Si
n[e + f*x])^2) - (a*(3*A*d + B*(c + 4*d))*Cos[e + f*x])/(4*d*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e
+ f*x]))
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Rubi [A] time = 0.370379, antiderivative size = 192, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used =
{2980, 2772, 2773, 208} \[ -\frac{\sqrt{a} (3 A d+B (c+4 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 d^{3/2} f (c+d)^{5/2}}-\frac{a (3 A d+B (c+4 d)) \cos (e+f x)}{4 d f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac{a (B c-A d) \cos (e+f x)}{2 d f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^2} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^3,x]
[Out]
-(Sqrt[a]*(3*A*d + B*(c + 4*d))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])]
)/(4*d^(3/2)*(c + d)^(5/2)*f) + (a*(B*c - A*d)*Cos[e + f*x])/(2*d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Si
n[e + f*x])^2) - (a*(3*A*d + B*(c + 4*d))*Cos[e + f*x])/(4*d*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e
+ f*x]))
Rule 2980
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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[n, -1]
Rule 2772
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
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{\sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+-\frac{(-3 a A d-B (a c+4 a d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^2} \, dx}{4 d (a c+a d)}\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac{a (3 A d+B (c+4 d)) \cos (e+f x)}{4 d (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac{(3 A d+B (c+4 d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 d (c+d)^2}\\ &=\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac{a (3 A d+B (c+4 d)) \cos (e+f x)}{4 d (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{(a (3 A d+B (c+4 d))) \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 )}{4 d (c+d)^2 f}\\ &=-\frac{\sqrt{a} (3 A d+B (c+4 d)) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{4 d^{3/2} (c+d)^{5/2} f}+\frac{a (B c-A d) \cos (e+f x)}{2 d (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac{a (3 A d+B (c+4 d)) \cos (e+f x)}{4 d (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end{align*}
Mathematica [C] time = 10.064, size = 967, normalized size = 5.04 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) \sqrt{a (\sin (e+f x)+1)} \left (\frac{(3 A d+B (c+4 d)) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right ) \left ((-1+i) x \cos (e)+(1+i) x \sin (e)+\frac{\text{RootSum}\left [d e^{2 i e} \text{$\#$1}^4+2 i c e^{i e} \text{$\#$1}^2-d\& ,\frac{-\sqrt{d} \sqrt{c+d} e^{i e} f x \text{$\#$1}^3-2 i \sqrt{d} \sqrt{c+d} e^{i e} \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right ) \text{$\#$1}^3+\frac{(1-i) c f x \text{$\#$1}^2}{\sqrt{e^{-i e}}}+\frac{(2+2 i) c \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right ) \text{$\#$1}^2}{\sqrt{e^{-i e}}}-i \sqrt{d} \sqrt{c+d} f x \text{$\#$1}+2 \sqrt{d} \sqrt{c+d} \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right ) \text{$\#$1}+(1+i) d \sqrt{e^{-i e}} f x-(2-2 i) d \sqrt{e^{-i e}} \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right )}{d-i c e^{i e} \text{$\#$1}^2}\& \right ] (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}{4 f}\right )}{(c+d)^{5/2} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}+\frac{(3 A d+B (c+4 d)) \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right ) \left ((1-i) x \cos (e)-(1+i) x \sin (e)+\frac{\text{RootSum}\left [d e^{2 i e} \text{$\#$1}^4+2 i c e^{i e} \text{$\#$1}^2-d\& ,\frac{-i \sqrt{d} \sqrt{c+d} e^{i e} f x \text{$\#$1}^3+2 \sqrt{d} \sqrt{c+d} e^{i e} \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right ) \text{$\#$1}^3-\frac{(1+i) c f x \text{$\#$1}^2}{\sqrt{e^{-i e}}}+\frac{(2-2 i) c \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right ) \text{$\#$1}^2}{\sqrt{e^{-i e}}}+\sqrt{d} \sqrt{c+d} f x \text{$\#$1}+2 i \sqrt{d} \sqrt{c+d} \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right ) \text{$\#$1}+(1-i) d \sqrt{e^{-i e}} f x+(2+2 i) d \sqrt{e^{-i e}} \log \left (e^{\frac{i f x}{2}}-\text{$\#$1}\right )}{d-i c e^{i e} \text{$\#$1}^2}\& \right ] \sqrt{\cos (e)-i \sin (e)} (-i \cos (e)+\sin (e)-1)}{4 f}\right )}{(c+d)^{5/2} (\cos (e)+i (\sin (e)-1)) \sqrt{\cos (e)-i \sin (e)}}-\frac{(2-2 i) \sqrt{d} (3 A d+B (c+4 d)) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d)^2 f (c+d \sin (e+f x))}-\frac{(4-4 i) \sqrt{d} (A d-B c) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d) f (c+d \sin (e+f x))^2}\right )}{d^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^3,x]
[Out]
((1/16 + I/16)*Sqrt[a*(1 + Sin[e + f*x])]*(((3*A*d + B*(c + 4*d))*(Cos[e/2] + I*Sin[e/2])*((-1 + I)*x*Cos[e] +
(RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 + I)*d*Sqrt[E^((-I)*e)]*f*x - (2 - 2*I)*d*Sqr
t[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] - I*Sqrt[d]*Sqrt[c + d]*f*x*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[E^((I/2)*f*x)
- #1]*#1 + ((1 - I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 + 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)
] - Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 - (2*I)*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #1]*#1^3)/(d
- I*c*E^(I*e)*#1^2) & ]*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]])/(4*f) + (1 + I)*x*Sin[e]))/((c + d
)^(5/2)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((3*A*d + B*(c + 4*d))*(Cos[e/2] + I*Sin[e/2])*(
(1 - I)*x*Cos[e] - (1 + I)*x*Sin[e] + (RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 - I)*d*S
qrt[E^((-I)*e)]*f*x + (2 + 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] + Sqrt[d]*Sqrt[c + d]*f*x*#1 + (2*I
)*Sqrt[d]*Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 - ((1 + I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 - 2*I)*c*Log[E^
((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - I*Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 + 2*Sqrt[d]*Sqrt[c + d]*E^(I
*e)*Log[E^((I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I*e)*#1^2) & ]*Sqrt[Cos[e] - I*Sin[e]]*(-1 - I*Cos[e] + Sin[e]))
/(4*f)))/((c + d)^(5/2)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((4 - 4*I)*Sqrt[d]*(-(B*c) + A*d
)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*f*(c + d*Sin[e + f*x])^2) - ((2 - 2*I)*Sqrt[d]*(3*A*d + B*(c
+ 4*d))*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)^2*f*(c + d*Sin[e + f*x]))))/(d^(3/2)*(Cos[(e + f*x)/2
] + Sin[(e + f*x)/2]))
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Maple [B] time = 2.075, size = 628, normalized size = 3.3 \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))^(1/2)/(c+d*sin(f*x+e))^3,x)
[Out]
1/4/a*(-2*sin(f*x+e)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d*(3*A*d+B*c+4*B*d)+arctanh((
a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(3*A*d+B*c+4*B*d)*cos(f*x+e)^2+3*A*(a-a*sin(f*x+e))^(3/2)
*(a*(c+d)*d)^(1/2)*d^2-3*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d-3*A*arctanh((a-a*si
n(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^3+B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c*d+4*B*(a-a*sin(f*x
+e))^(3/2)*(a*(c+d)*d)^(1/2)*d^2-a^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*B*c^3-4*B*arctanh((
a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d-B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))
*a^2*c*d^2-4*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^3-5*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+
d)*d)^(1/2)*a*c*d-5*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^2+B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2
)*a*c^2-3*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d-4*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^2)
*(-a*(-1+sin(f*x+e)))^(1/2)*(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))^2/(c+d)^2/d/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))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] time = 16.4405, size = 3954, normalized size = 20.59 \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))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
[-1/16*((B*c^3 + 3*(A + 2*B)*c^2*d + 3*(2*A + 3*B)*c*d^2 + (3*A + 4*B)*d^3 - (B*c*d^2 + (3*A + 4*B)*d^3)*cos(f
*x + e)^3 - (2*B*c^2*d + 3*(2*A + 3*B)*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e)^2 + (B*c^3 + (3*A + 4*B)*c^2*d +
B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e) + (B*c^3 + 3*(A + 2*B)*c^2*d + 3*(2*A + 3*B)*c*d^2 + (3*A + 4*B)*d^3 -
(B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e)^2 + 2*(B*c^2*d + (3*A + 4*B)*c*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt
(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(
c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c
*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2
+ 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*c
os(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*(B*c^2 - (5*
A + B)*c*d + (A + 4*B)*d^2 - (B*c*d + (3*A + 4*B)*d^2)*cos(f*x + e)^2 + (B*c^2 - (5*A + 2*B)*c*d - 2*A*d^2)*co
s(f*x + e) - (B*c^2 - (5*A + B)*c*d + (A + 4*B)*d^2 + (B*c*d + (3*A + 4*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sq
rt(a*sin(f*x + e) + a))/((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 + 4*c*d^4 + d^5)*
f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2
*d^3 + 4*c*d^4 + d^5)*f + ((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d^3 + c*d^4)*f*cos(
f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*f)*sin(f*x + e)), 1/8*((B*c^3 + 3*(A + 2*B)*c^2*d +
3*(2*A + 3*B)*c*d^2 + (3*A + 4*B)*d^3 - (B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e)^3 - (2*B*c^2*d + 3*(2*A + 3*
B)*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e)^2 + (B*c^3 + (3*A + 4*B)*c^2*d + B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x +
e) + (B*c^3 + 3*(A + 2*B)*c^2*d + 3*(2*A + 3*B)*c*d^2 + (3*A + 4*B)*d^3 - (B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x
+ e)^2 + 2*(B*c^2*d + (3*A + 4*B)*c*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*s
in(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(B*c^2 - (5*A + B)*c*d
+ (A + 4*B)*d^2 - (B*c*d + (3*A + 4*B)*d^2)*cos(f*x + e)^2 + (B*c^2 - (5*A + 2*B)*c*d - 2*A*d^2)*cos(f*x + e)
- (B*c^2 - (5*A + B)*c*d + (A + 4*B)*d^2 + (B*c*d + (3*A + 4*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*
x + e) + a))/((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c^2*d^3 + 4*c*d^4 + d^5)*f*cos(f*x +
e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e) - (c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*
d^4 + d^5)*f + ((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^2 - 2*(c^3*d^2 + 2*c^2*d^3 + c*d^4)*f*cos(f*x + e) -
(c^4*d + 4*c^3*d^2 + 6*c^2*d^3 + 4*c*d^4 + d^5)*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))**(1/2)/(c+d*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")
[Out]
Exception raised: TypeError