3.6.25 \(\int \frac {\sqrt {a+b x^3} (A+B x^3)}{x^{11/2}} \, dx\) [525]
Optimal. Leaf size=79 \[ -\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+\frac {2}{3} \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right ) \]
[Out]
-2/9*A*(b*x^3+a)^(3/2)/a/x^(9/2)+2/3*B*arctanh(x^(3/2)*b^(1/2)/(b*x^3+a)^(1/2))*b^(1/2)-2/3*B*(b*x^3+a)^(1/2)/
x^(3/2)
________________________________________________________________________________________
Rubi [A]
time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {462, 283, 335,
281, 223, 212} \begin {gather*} -\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}+\frac {2}{3} \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Sqrt[a + b*x^3]*(A + B*x^3))/x^(11/2),x]
[Out]
(-2*B*Sqrt[a + b*x^3])/(3*x^(3/2)) - (2*A*(a + b*x^3)^(3/2))/(9*a*x^(9/2)) + (2*Sqrt[b]*B*ArcTanh[(Sqrt[b]*x^(
3/2))/Sqrt[a + b*x^3]])/3
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 223
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 281
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 283
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 335
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 462
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, 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] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))
Rubi steps
\begin {align*} \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+B \int \frac {\sqrt {a+b x^3}}{x^{5/2}} \, dx\\ &=-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+(b B) \int \frac {\sqrt {x}}{\sqrt {a+b x^3}} \, dx\\ &=-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+(2 b B) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^6}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+\frac {1}{3} (2 b B) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^{3/2}\right )\\ &=-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+\frac {1}{3} (2 b B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a+b x^3}}\right )\\ &=-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+\frac {2}{3} \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right )\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 0.25, size = 74, normalized size = 0.94 \begin {gather*} -\frac {2 \sqrt {a+b x^3} \left (a A+A b x^3+3 a B x^3\right )}{9 a x^{9/2}}+\frac {2}{3} \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[a + b*x^3]*(A + B*x^3))/x^(11/2),x]
[Out]
(-2*Sqrt[a + b*x^3]*(a*A + A*b*x^3 + 3*a*B*x^3))/(9*a*x^(9/2)) + (2*Sqrt[b]*B*ArcTanh[Sqrt[a + b*x^3]/(Sqrt[b]
*x^(3/2))])/3
________________________________________________________________________________________
Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.42, size = 3759, normalized size = 47.58
| | |
| method |
result |
size |
| | |
| risch |
\(\text {Expression too large to display}\) |
\(1040\) |
| elliptic |
\(\text {Expression too large to display}\) |
\(1051\) |
| default |
\(\text {Expression too large to display}\) |
\(3759\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x^3+A)*(b*x^3+a)^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)
[Out]
-2/9*(b*x^3+a)^(1/2)/x^(9/2)/b*(18*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+
I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a*b^
2*x^7-3*A*(1/b^2*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*(b*x^3+a))^(1/2)*b^2*x^3+36*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*Elli
pticPi((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2
)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*a*b*x^6-3*A*(1/b^2*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*
(b*x^3+a))^(1/2)*a*b+18*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/
(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(2/3)*
a*x^5+I*A*3^(1/2)*(1/b^2*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*(b*x^3+a))^(1/2)*b^2*x^3-18*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1
/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((
I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a*b^2*x^7+3*I*B*3^(1/2)*(1/b^2*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*(b*x^3+a))^(1/2)*a*b*x^3-18*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(-1+I
*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2)
)/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(2/3)*a*x^5-36*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2
)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3)
)^(1/2))*(-a*b^2)^(1/3)*a*b*x^6-18*B*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))
/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a*b^2*x^7+18*B
*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))
/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),(-1+I
*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a*b^2*x^7+36*B*(-(I*
3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x
+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+
3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a*b^2)^(1/3)*a*b*x^6-36*B*(-(I*3^(1/2)-3)*x*b/(-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))/(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))/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)
*EllipticPi((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3
^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/...
________________________________________________________________________________________
Maxima [A]
time = 0.50, size = 81, normalized size = 1.03 \begin {gather*} -\frac {1}{3} \, {\left (\sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}\right ) + \frac {2 \, \sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}\right )} B - \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A}{9 \, a x^{\frac {9}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^(11/2),x, algorithm="maxima")
[Out]
-1/3*(sqrt(b)*log(-(sqrt(b) - sqrt(b*x^3 + a)/x^(3/2))/(sqrt(b) + sqrt(b*x^3 + a)/x^(3/2))) + 2*sqrt(b*x^3 + a
)/x^(3/2))*B - 2/9*(b*x^3 + a)^(3/2)*A/(a*x^(9/2))
________________________________________________________________________________________
Fricas [A]
time = 1.97, size = 180, normalized size = 2.28 \begin {gather*} \left [\frac {3 \, B a \sqrt {b} x^{5} \log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b} \sqrt {x} - a^{2}\right ) - 4 \, {\left ({\left (3 \, B a + A b\right )} x^{3} + A a\right )} \sqrt {b x^{3} + a} \sqrt {x}}{18 \, a x^{5}}, -\frac {3 \, B a \sqrt {-b} x^{5} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b} x^{\frac {3}{2}}}{2 \, b x^{3} + a}\right ) + 2 \, {\left ({\left (3 \, B a + A b\right )} x^{3} + A a\right )} \sqrt {b x^{3} + a} \sqrt {x}}{9 \, a x^{5}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^(11/2),x, algorithm="fricas")
[Out]
[1/18*(3*B*a*sqrt(b)*x^5*log(-8*b^2*x^6 - 8*a*b*x^3 - 4*(2*b*x^4 + a*x)*sqrt(b*x^3 + a)*sqrt(b)*sqrt(x) - a^2)
- 4*((3*B*a + A*b)*x^3 + A*a)*sqrt(b*x^3 + a)*sqrt(x))/(a*x^5), -1/9*(3*B*a*sqrt(-b)*x^5*arctan(2*sqrt(b*x^3
+ a)*sqrt(-b)*x^(3/2)/(2*b*x^3 + a)) + 2*((3*B*a + A*b)*x^3 + A*a)*sqrt(b*x^3 + a)*sqrt(x))/(a*x^5)]
________________________________________________________________________________________
Sympy [A]
time = 32.28, size = 131, normalized size = 1.66 \begin {gather*} - \frac {2 A \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{9 x^{3}} - \frac {2 A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}}{9 a} - \frac {2 B \sqrt {a}}{3 x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {2 B \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3} - \frac {2 B b x^{\frac {3}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x**3+A)*(b*x**3+a)**(1/2)/x**(11/2),x)
[Out]
-2*A*sqrt(b)*sqrt(a/(b*x**3) + 1)/(9*x**3) - 2*A*b**(3/2)*sqrt(a/(b*x**3) + 1)/(9*a) - 2*B*sqrt(a)/(3*x**(3/2)
*sqrt(1 + b*x**3/a)) + 2*B*sqrt(b)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/3 - 2*B*b*x**(3/2)/(3*sqrt(a)*sqrt(1 + b*x*
*3/a))
________________________________________________________________________________________
Giac [A]
time = 0.82, size = 109, normalized size = 1.38 \begin {gather*} -\frac {2 \, B b \arctan \left (\frac {\sqrt {b + \frac {a}{x^{3}}}}{\sqrt {-b}}\right )}{3 \, \sqrt {-b}} + \frac {2 \, {\left (3 \, B a b \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 3 \, B a \sqrt {-b} \sqrt {b} + A \sqrt {-b} b^{\frac {3}{2}}\right )}}{9 \, a \sqrt {-b}} - \frac {2 \, {\left (3 \, B a^{3} \sqrt {b + \frac {a}{x^{3}}} + A a^{2} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {3}{2}}\right )}}{9 \, a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^(11/2),x, algorithm="giac")
[Out]
-2/3*B*b*arctan(sqrt(b + a/x^3)/sqrt(-b))/sqrt(-b) + 2/9*(3*B*a*b*arctan(sqrt(b)/sqrt(-b)) + 3*B*a*sqrt(-b)*sq
rt(b) + A*sqrt(-b)*b^(3/2))/(a*sqrt(-b)) - 2/9*(3*B*a^3*sqrt(b + a/x^3) + A*a^2*(b + a/x^3)^(3/2))/a^3
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^3+A\right )\,\sqrt {b\,x^3+a}}{x^{11/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*x^3)*(a + b*x^3)^(1/2))/x^(11/2),x)
[Out]
int(((A + B*x^3)*(a + b*x^3)^(1/2))/x^(11/2), x)
________________________________________________________________________________________