Integrand size = 24, antiderivative size = 79 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=-\frac {2 B \sqrt {a+b x^3}}{3 x^{3/2}}-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}+\frac {2}{3} \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right ) \] Output:
-2/3*B*(b*x^3+a)^(1/2)/x^(3/2)-2/9*A*(b*x^3+a)^(3/2)/a/x^(9/2)+2/3*b^(1/2) *B*arctanh(b^(1/2)*x^(3/2)/(b*x^3+a)^(1/2))
Time = 0.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=-\frac {2 \sqrt {a+b x^3} \left (a A+A b x^3+3 a B x^3\right )}{9 a x^{9/2}}+\frac {2}{3} \sqrt {b} B \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right ) \] Input:
Integrate[(Sqrt[a + b*x^3]*(A + B*x^3))/x^(11/2),x]
Output:
(-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*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]])/3
Time = 0.38 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {953, 809, 851, 807, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle B \int \frac {\sqrt {b x^3+a}}{x^{5/2}}dx-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle B \left (b \int \frac {\sqrt {x}}{\sqrt {b x^3+a}}dx-\frac {2 \sqrt {a+b x^3}}{3 x^{3/2}}\right )-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle B \left (2 b \int \frac {x}{\sqrt {b x^3+a}}d\sqrt {x}-\frac {2 \sqrt {a+b x^3}}{3 x^{3/2}}\right )-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle B \left (\frac {2}{3} b \int \frac {1}{\sqrt {a+b x}}dx^{3/2}-\frac {2 \sqrt {a+b x^3}}{3 x^{3/2}}\right )-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle B \left (\frac {2}{3} b \int \frac {1}{1-b x}d\frac {x^{3/2}}{\sqrt {a+b x}}-\frac {2 \sqrt {a+b x^3}}{3 x^{3/2}}\right )-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle B \left (\frac {2}{3} \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x}}\right )-\frac {2 \sqrt {a+b x^3}}{3 x^{3/2}}\right )-\frac {2 A \left (a+b x^3\right )^{3/2}}{9 a x^{9/2}}\) |
Input:
Int[(Sqrt[a + b*x^3]*(A + B*x^3))/x^(11/2),x]
Output:
(-2*A*(a + b*x^3)^(3/2))/(9*a*x^(9/2)) + B*((-2*Sqrt[a + b*x^3])/(3*x^(3/2 )) + (2*Sqrt[b]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a + b*x]])/3)
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
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] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(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] && IntB inomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[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]
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] + Simp[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] && G tQ[m + n, -1]))
Time = 1.57 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06
method | result | size |
risch | \(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (A b \,x^{3}+3 B a \,x^{3}+A a \right )}{9 x^{\frac {9}{2}} a}+\frac {2 B \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b}}\right ) \sqrt {x \left (b \,x^{3}+a \right )}}{3 \sqrt {x}\, \sqrt {b \,x^{3}+a}}\) | \(84\) |
default | \(-\frac {2 \sqrt {b \,x^{3}+a}\, \left (-3 B \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b}}\right ) \sqrt {b}\, a \,x^{5}+A \sqrt {x \left (b \,x^{3}+a \right )}\, b \,x^{3}+3 B \sqrt {x \left (b \,x^{3}+a \right )}\, a \,x^{3}+A \sqrt {x \left (b \,x^{3}+a \right )}\, a \right )}{9 x^{\frac {9}{2}} \sqrt {x \left (b \,x^{3}+a \right )}\, a}\) | \(108\) |
elliptic | \(\text {Expression too large to display}\) | \(1051\) |
Input:
int((b*x^3+a)^(1/2)*(B*x^3+A)/x^(11/2),x,method=_RETURNVERBOSE)
Output:
-2/9*(b*x^3+a)^(1/2)*(A*b*x^3+3*B*a*x^3+A*a)/x^(9/2)/a+2/3*B*b^(1/2)*arcta nh(1/x^2*(x*(b*x^3+a))^(1/2)/b^(1/2))*(x*(b*x^3+a))^(1/2)/x^(1/2)/(b*x^3+a )^(1/2)
Time = 0.30 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.28 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=\left [\frac {3 \, B a \sqrt {b} x^{5} \log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b} \sqrt {x} - a^{2}\right ) - 4 \, {\left ({\left (3 \, B a + A b\right )} x^{3} + A a\right )} \sqrt {b x^{3} + a} \sqrt {x}}{18 \, a x^{5}}, -\frac {3 \, B a \sqrt {-b} x^{5} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b} x^{\frac {3}{2}}}{2 \, b x^{3} + a}\right ) + 2 \, {\left ({\left (3 \, B a + A b\right )} x^{3} + A a\right )} \sqrt {b x^{3} + a} \sqrt {x}}{9 \, a x^{5}}\right ] \] Input:
integrate((b*x^3+a)^(1/2)*(B*x^3+A)/x^(11/2),x, algorithm="fricas")
Output:
[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)*sq rt(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)]
Time = 27.57 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.66 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=- \frac {2 A \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{9 x^{3}} - \frac {2 A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}}{9 a} - \frac {2 B \sqrt {a}}{3 x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {2 B \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3} - \frac {2 B b x^{\frac {3}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} \] Input:
integrate((b*x**3+a)**(1/2)*(B*x**3+A)/x**(11/2),x)
Output:
-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)*asi nh(sqrt(b)*x**(3/2)/sqrt(a))/3 - 2*B*b*x**(3/2)/(3*sqrt(a)*sqrt(1 + b*x**3 /a))
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=-\frac {1}{3} \, {\left (\sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}}\right ) + \frac {2 \, \sqrt {b x^{3} + a}}{x^{\frac {3}{2}}}\right )} B - \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A}{9 \, a x^{\frac {9}{2}}} \] Input:
integrate((b*x^3+a)^(1/2)*(B*x^3+A)/x^(11/2),x, algorithm="maxima")
Output:
-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))
Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=-\frac {2 \, B b \arctan \left (\frac {\sqrt {b + \frac {a}{x^{3}}}}{\sqrt {-b}}\right )}{3 \, \sqrt {-b}} + \frac {2 \, {\left (3 \, B a b \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 3 \, B a \sqrt {-b} \sqrt {b} + A \sqrt {-b} b^{\frac {3}{2}}\right )}}{9 \, a \sqrt {-b}} - \frac {2 \, {\left (3 \, B a^{3} \sqrt {b + \frac {a}{x^{3}}} + A a^{2} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {3}{2}}\right )}}{9 \, a^{3}} \] Input:
integrate((b*x^3+a)^(1/2)*(B*x^3+A)/x^(11/2),x, algorithm="giac")
Output:
-2/3*B*b*arctan(sqrt(b + a/x^3)/sqrt(-b))/sqrt(-b) + 2/9*(3*B*a*b*arctan(s qrt(b)/sqrt(-b)) + 3*B*a*sqrt(-b)*sqrt(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
Timed out. \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,\sqrt {b\,x^3+a}}{x^{11/2}} \,d x \] Input:
int(((A + B*x^3)*(a + b*x^3)^(1/2))/x^(11/2),x)
Output:
int(((A + B*x^3)*(a + b*x^3)^(1/2))/x^(11/2), x)
Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^{11/2}} \, dx=\frac {-2 \sqrt {b \,x^{3}+a}\, a -8 \sqrt {b \,x^{3}+a}\, b \,x^{3}-3 \sqrt {x}\, \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) b \,x^{4}+3 \sqrt {x}\, \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) b \,x^{4}}{9 \sqrt {x}\, x^{4}} \] Input:
int((b*x^3+a)^(1/2)*(B*x^3+A)/x^(11/2),x)
Output:
( - 2*sqrt(a + b*x**3)*a - 8*sqrt(a + b*x**3)*b*x**3 - 3*sqrt(x)*sqrt(b)*l og(sqrt(a + b*x**3) - sqrt(x)*sqrt(b)*x)*b*x**4 + 3*sqrt(x)*sqrt(b)*log(sq rt(a + b*x**3) + sqrt(x)*sqrt(b)*x)*b*x**4)/(9*sqrt(x)*x**4)