3.554 \(\int \frac {\sqrt {e x} (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\)
Optimal. Leaf size=85 \[ \frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}+\frac {2 B \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{3/2}} \]
[Out]
2/3*B*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))*e^(1/2)/b^(3/2)+2/3*(A*b-B*a)*(e*x)^(3/2)/a/b/e/(b*
x^3+a)^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used =
{452, 329, 275, 217, 206} \[ \frac {2 (e x)^{3/2} (A b-a B)}{3 a b e \sqrt {a+b x^3}}+\frac {2 B \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[e*x]*(A + B*x^3))/(a + b*x^3)^(3/2),x]
[Out]
(2*(A*b - a*B)*(e*x)^(3/2))/(3*a*b*e*Sqrt[a + b*x^3]) + (2*B*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sq
rt[a + b*x^3])])/(3*b^(3/2))
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 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 275
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Rule 329
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 452
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[((b*c - a*d)
*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*(m + 1)), x] + Dist[d/b, Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /;
FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && NeQ[m, -1]
Rubi steps
\begin {align*} \int \frac {\sqrt {e x} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac {2 (A b-a B) (e x)^{3/2}}{3 a b e \sqrt {a+b x^3}}+\frac {B \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{b}\\ &=\frac {2 (A b-a B) (e x)^{3/2}}{3 a b e \sqrt {a+b x^3}}+\frac {(2 B) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{b e}\\ &=\frac {2 (A b-a B) (e x)^{3/2}}{3 a b e \sqrt {a+b x^3}}+\frac {(2 B) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{3 b e}\\ &=\frac {2 (A b-a B) (e x)^{3/2}}{3 a b e \sqrt {a+b x^3}}+\frac {(2 B) \operatorname {Subst}\left (\int \frac {1}{1-\frac {b x^2}{e^3}} \, dx,x,\frac {(e x)^{3/2}}{\sqrt {a+b x^3}}\right )}{3 b e}\\ &=\frac {2 (A b-a B) (e x)^{3/2}}{3 a b e \sqrt {a+b x^3}}+\frac {2 B \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{3/2}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.10, size = 93, normalized size = 1.09 \[ \frac {2 \sqrt {e x} \left (a^{3/2} B \sqrt {\frac {b x^3}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )+\sqrt {b} x^{3/2} (A b-a B)\right )}{3 a b^{3/2} \sqrt {x} \sqrt {a+b x^3}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[e*x]*(A + B*x^3))/(a + b*x^3)^(3/2),x]
[Out]
(2*Sqrt[e*x]*(Sqrt[b]*(A*b - a*B)*x^(3/2) + a^(3/2)*B*Sqrt[1 + (b*x^3)/a]*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))
/(3*a*b^(3/2)*Sqrt[x]*Sqrt[a + b*x^3])
________________________________________________________________________________________
fricas [A] time = 1.30, size = 234, normalized size = 2.75 \[ \left [-\frac {4 \, \sqrt {b x^{3} + a} {\left (B a - A b\right )} \sqrt {e x} x - {\left (B a b x^{3} + B a^{2}\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right )}{6 \, {\left (a b^{2} x^{3} + a^{2} b\right )}}, -\frac {2 \, \sqrt {b x^{3} + a} {\left (B a - A b\right )} \sqrt {e x} x + {\left (B a b x^{3} + B a^{2}\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right )}{3 \, {\left (a b^{2} x^{3} + a^{2} b\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^3+A)*(e*x)^(1/2)/(b*x^3+a)^(3/2),x, algorithm="fricas")
[Out]
[-1/6*(4*sqrt(b*x^3 + a)*(B*a - A*b)*sqrt(e*x)*x - (B*a*b*x^3 + B*a^2)*sqrt(e/b)*log(-8*b^2*e*x^6 - 8*a*b*e*x^
3 - a^2*e - 4*(2*b^2*x^4 + a*b*x)*sqrt(b*x^3 + a)*sqrt(e*x)*sqrt(e/b)))/(a*b^2*x^3 + a^2*b), -1/3*(2*sqrt(b*x^
3 + a)*(B*a - A*b)*sqrt(e*x)*x + (B*a*b*x^3 + B*a^2)*sqrt(-e/b)*arctan(2*sqrt(b*x^3 + a)*sqrt(e*x)*b*x*sqrt(-e
/b)/(2*b*e*x^3 + a*e)))/(a*b^2*x^3 + a^2*b)]
________________________________________________________________________________________
giac [A] time = 0.62, size = 75, normalized size = 0.88 \[ -\frac {2 \, B e^{\frac {1}{2}} \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} e^{2} + \sqrt {b x^{3} e^{4} + a e^{4}} \right |}\right )}{3 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (B a e - A b e\right )} x^{\frac {3}{2}} e^{\frac {3}{2}}}{3 \, \sqrt {b x^{3} e^{4} + a e^{4}} a b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^3+A)*(e*x)^(1/2)/(b*x^3+a)^(3/2),x, algorithm="giac")
[Out]
-2/3*B*e^(1/2)*log(abs(-sqrt(b)*x^(3/2)*e^2 + sqrt(b*x^3*e^4 + a*e^4)))/b^(3/2) - 2/3*(B*a*e - A*b*e)*x^(3/2)*
e^(3/2)/(sqrt(b*x^3*e^4 + a*e^4)*a*b)
________________________________________________________________________________________
maple [C] time = 1.00, size = 3654, normalized size = 42.99 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x^3+A)*(e*x)^(1/2)/(b*x^3+a)^(3/2),x)
[Out]
2/3*(e*x)^(1/2)/(b*x^3+a)^(1/2)/b^3*(-6*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(
1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(
-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(
I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*
x*(b*x^3+a))^(1/2)*x^2*a*b^2-12*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I
*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^
(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(I*3^(1/
2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3
^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*x*a*b+12*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/
(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/
3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Ellipti
cF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(
I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*x*a*b-6*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1
)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(
1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Ellip
ticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))
/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(2/3)*a+6*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/
(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/
3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Ellipti
cPi((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*
(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*x^2*a*b^2+I*A*3^(1/2)*(1/b^2*e*x*(-b*x
+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3
)))^(1/2)*x^2*b^3+6*B*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3
)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(
1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1
/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*x^2*a*b^2-6
*B*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1
/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)
/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),(I*3^(
1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*x
^2*a*b^2-12*B*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^
2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a
*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*
b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*x*a*
b+12*B*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3
)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(
1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(
1/3)))^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x
*(b*x^3+a))^(1/2)*x*a*b-I*B*3^(1/2)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^
(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^2*a*b^2+6*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3
^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-
a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/
2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3
^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(2/3)*a+6*B*(-a*b^2
)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^
2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/
2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((
I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*a-6*B*(-a*b^2)^(2/3)*(-(I
*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1
+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+
(-a*b^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),(I*3^(1/2)-1)
/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(e*x*(b*x^3+a))^(1/2)*a-3*A*(1
/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x
-(-a*b^2)^(1/3)))^(1/2)*x^2*b^3+3*B*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^
(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^2*a*b^2)/x/(I*3^(1/2)-3)/(1/b^2*e*x*(-b*x+(-a*
b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(
1/2)/a
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} \sqrt {e x}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^3+A)*(e*x)^(1/2)/(b*x^3+a)^(3/2),x, algorithm="maxima")
[Out]
integrate((B*x^3 + A)*sqrt(e*x)/(b*x^3 + a)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^3+A\right )\,\sqrt {e\,x}}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*x^3)*(e*x)^(1/2))/(a + b*x^3)^(3/2),x)
[Out]
int(((A + B*x^3)*(e*x)^(1/2))/(a + b*x^3)^(3/2), x)
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sympy [A] time = 33.62, size = 95, normalized size = 1.12 \[ \frac {2 A \sqrt {e} x^{\frac {3}{2}}}{3 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + B \left (\frac {2 \sqrt {e} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 b^{\frac {3}{2}}} - \frac {2 \sqrt {e} x^{\frac {3}{2}}}{3 \sqrt {a} b \sqrt {1 + \frac {b x^{3}}{a}}}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x**3+A)*(e*x)**(1/2)/(b*x**3+a)**(3/2),x)
[Out]
2*A*sqrt(e)*x**(3/2)/(3*a**(3/2)*sqrt(1 + b*x**3/a)) + B*(2*sqrt(e)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*b**(3/2
)) - 2*sqrt(e)*x**(3/2)/(3*sqrt(a)*b*sqrt(1 + b*x**3/a)))
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