Integrand size = 22, antiderivative size = 100 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\frac {2}{3} b B \sqrt {a+b x^3}-\frac {(3 A b+4 a B) \sqrt {a+b x^3}}{12 x^3}-\frac {A \left (a+b x^3\right )^{3/2}}{6 x^6}-\frac {b (A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}} \] Output:
2/3*b*B*(b*x^3+a)^(1/2)-1/12*(3*A*b+4*B*a)*(b*x^3+a)^(1/2)/x^3-1/6*A*(b*x^ 3+a)^(3/2)/x^6-1/4*b*(A*b+4*B*a)*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(1/2)
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\frac {\sqrt {a+b x^3} \left (-2 a A-5 A b x^3-4 a B x^3+8 b B x^6\right )}{12 x^6}-\frac {b (A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}} \] Input:
Integrate[((a + b*x^3)^(3/2)*(A + B*x^3))/x^7,x]
Output:
(Sqrt[a + b*x^3]*(-2*a*A - 5*A*b*x^3 - 4*a*B*x^3 + 8*b*B*x^6))/(12*x^6) - (b*(A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*Sqrt[a])
Time = 0.36 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 87, 51, 60, 73, 221}
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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {\left (b x^3+a\right )^{3/2} \left (B x^3+A\right )}{x^9}dx^3\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{3} \left (\frac {(4 a B+A b) \int \frac {\left (b x^3+a\right )^{3/2}}{x^6}dx^3}{4 a}-\frac {A \left (a+b x^3\right )^{5/2}}{2 a x^6}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} \left (\frac {(4 a B+A b) \left (\frac {3}{2} b \int \frac {\sqrt {b x^3+a}}{x^3}dx^3-\frac {\left (a+b x^3\right )^{3/2}}{x^3}\right )}{4 a}-\frac {A \left (a+b x^3\right )^{5/2}}{2 a x^6}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (\frac {(4 a B+A b) \left (\frac {3}{2} b \left (a \int \frac {1}{x^3 \sqrt {b x^3+a}}dx^3+2 \sqrt {a+b x^3}\right )-\frac {\left (a+b x^3\right )^{3/2}}{x^3}\right )}{4 a}-\frac {A \left (a+b x^3\right )^{5/2}}{2 a x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {(4 a B+A b) \left (\frac {3}{2} b \left (\frac {2 a \int \frac {1}{\frac {x^6}{b}-\frac {a}{b}}d\sqrt {b x^3+a}}{b}+2 \sqrt {a+b x^3}\right )-\frac {\left (a+b x^3\right )^{3/2}}{x^3}\right )}{4 a}-\frac {A \left (a+b x^3\right )^{5/2}}{2 a x^6}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {(4 a B+A b) \left (\frac {3}{2} b \left (2 \sqrt {a+b x^3}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\right )-\frac {\left (a+b x^3\right )^{3/2}}{x^3}\right )}{4 a}-\frac {A \left (a+b x^3\right )^{5/2}}{2 a x^6}\right )\) |
Input:
Int[((a + b*x^3)^(3/2)*(A + B*x^3))/x^7,x]
Output:
(-1/2*(A*(a + b*x^3)^(5/2))/(a*x^6) + ((A*b + 4*a*B)*(-((a + b*x^3)^(3/2)/ x^3) + (3*b*(2*Sqrt[a + b*x^3] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a] ]))/2))/(4*a))/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.88 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(-\frac {b \,x^{6} \left (A b +4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )+\frac {5 \sqrt {b \,x^{3}+a}\, \left (\frac {2 \left (2 B \,x^{3}+A \right ) a^{\frac {3}{2}}}{5}+b \,x^{3} \sqrt {a}\, \left (-\frac {8 B \,x^{3}}{5}+A \right )\right )}{3}}{4 \sqrt {a}\, x^{6}}\) | \(78\) |
risch | \(-\frac {\sqrt {b \,x^{3}+a}\, \left (5 A b \,x^{3}+4 B a \,x^{3}+2 A a \right )}{12 x^{6}}+\frac {b \left (-\frac {2 \left (3 A b +12 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \sqrt {a}}+\frac {16 B \sqrt {b \,x^{3}+a}}{3}\right )}{8}\) | \(79\) |
elliptic | \(-\frac {A a \sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {\left (\frac {5 A b}{4}+B a \right ) \sqrt {b \,x^{3}+a}}{3 x^{3}}+\frac {2 b B \sqrt {b \,x^{3}+a}}{3}-\frac {2 \left (\frac {3}{8} b^{2} A +\frac {3}{2} a b B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \sqrt {a}}\) | \(84\) |
default | \(A \left (-\frac {a \sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {5 b \sqrt {b \,x^{3}+a}}{12 x^{3}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 \sqrt {a}}\right )+B \left (\frac {2 b \sqrt {b \,x^{3}+a}}{3}-\frac {a \sqrt {b \,x^{3}+a}}{3 x^{3}}-\sqrt {a}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )\right )\) | \(107\) |
Input:
int((b*x^3+a)^(3/2)*(B*x^3+A)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/4*(b*x^6*(A*b+4*B*a)*arctanh((b*x^3+a)^(1/2)/a^(1/2))+5/3*(b*x^3+a)^(1/ 2)*(2/5*(2*B*x^3+A)*a^(3/2)+b*x^3*a^(1/2)*(-8/5*B*x^3+A)))/a^(1/2)/x^6
Time = 0.12 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.88 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\left [\frac {3 \, {\left (4 \, B a b + A b^{2}\right )} \sqrt {a} x^{6} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (8 \, B a b x^{6} - {\left (4 \, B a^{2} + 5 \, A a b\right )} x^{3} - 2 \, A a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a x^{6}}, \frac {3 \, {\left (4 \, B a b + A b^{2}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{3} + a}}\right ) + {\left (8 \, B a b x^{6} - {\left (4 \, B a^{2} + 5 \, A a b\right )} x^{3} - 2 \, A a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a x^{6}}\right ] \] Input:
integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^7,x, algorithm="fricas")
Output:
[1/24*(3*(4*B*a*b + A*b^2)*sqrt(a)*x^6*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt (a) + 2*a)/x^3) + 2*(8*B*a*b*x^6 - (4*B*a^2 + 5*A*a*b)*x^3 - 2*A*a^2)*sqrt (b*x^3 + a))/(a*x^6), 1/12*(3*(4*B*a*b + A*b^2)*sqrt(-a)*x^6*arctan(sqrt(- a)/sqrt(b*x^3 + a)) + (8*B*a*b*x^6 - (4*B*a^2 + 5*A*a*b)*x^3 - 2*A*a^2)*sq rt(b*x^3 + a))/(a*x^6)]
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (90) = 180\).
Time = 46.20 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.43 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=- \frac {A a^{2}}{6 \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {A a \sqrt {b}}{4 x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}}{3 x^{\frac {3}{2}}} - \frac {A b^{\frac {3}{2}}}{12 x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 \sqrt {a}} - B \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )} - \frac {B a \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 x^{\frac {3}{2}}} + \frac {2 B a \sqrt {b}}{3 x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {2 B b^{\frac {3}{2}} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x^{3}} + 1}} \] Input:
integrate((b*x**3+a)**(3/2)*(B*x**3+A)/x**7,x)
Output:
-A*a**2/(6*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) - A*a*sqrt(b)/(4*x**(9/ 2)*sqrt(a/(b*x**3) + 1)) - A*b**(3/2)*sqrt(a/(b*x**3) + 1)/(3*x**(3/2)) - A*b**(3/2)/(12*x**(3/2)*sqrt(a/(b*x**3) + 1)) - A*b**2*asinh(sqrt(a)/(sqrt (b)*x**(3/2)))/(4*sqrt(a)) - B*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x**(3/2))) - B*a*sqrt(b)*sqrt(a/(b*x**3) + 1)/(3*x**(3/2)) + 2*B*a*sqrt(b)/(3*x**(3/ 2)*sqrt(a/(b*x**3) + 1)) + 2*B*b**(3/2)*x**(3/2)/(3*sqrt(a/(b*x**3) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (80) = 160\).
Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\frac {1}{24} \, {\left (\frac {3 \, b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 \, {\left (5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {b x^{3} + a} a b^{2}\right )}}{{\left (b x^{3} + a\right )}^{2} - 2 \, {\left (b x^{3} + a\right )} a + a^{2}}\right )} A + \frac {1}{6} \, {\left (3 \, \sqrt {a} b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + 4 \, \sqrt {b x^{3} + a} b - \frac {2 \, \sqrt {b x^{3} + a} a}{x^{3}}\right )} B \] Input:
integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^7,x, algorithm="maxima")
Output:
1/24*(3*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/s qrt(a) - 2*(5*(b*x^3 + a)^(3/2)*b^2 - 3*sqrt(b*x^3 + a)*a*b^2)/((b*x^3 + a )^2 - 2*(b*x^3 + a)*a + a^2))*A + 1/6*(3*sqrt(a)*b*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a))) + 4*sqrt(b*x^3 + a)*b - 2*sqrt(b*x^3 + a)*a/x^3)*B
Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\frac {8 \, \sqrt {b x^{3} + a} B b^{2} + \frac {3 \, {\left (4 \, B a b^{2} + A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x^{3} + a} B a^{2} b^{2} + 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b^{3} - 3 \, \sqrt {b x^{3} + a} A a b^{3}}{b^{2} x^{6}}}{12 \, b} \] Input:
integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^7,x, algorithm="giac")
Output:
1/12*(8*sqrt(b*x^3 + a)*B*b^2 + 3*(4*B*a*b^2 + A*b^3)*arctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) - (4*(b*x^3 + a)^(3/2)*B*a*b^2 - 4*sqrt(b*x^3 + a)*B *a^2*b^2 + 5*(b*x^3 + a)^(3/2)*A*b^3 - 3*sqrt(b*x^3 + a)*A*a*b^3)/(b^2*x^6 ))/b
Time = 1.58 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\frac {2\,B\,b\,\sqrt {b\,x^3+a}}{3}-\frac {\sqrt {b\,x^3+a}\,\left (4\,B\,a^3+5\,A\,b\,a^2\right )}{12\,a^2\,x^3}-\frac {A\,a\,\sqrt {b\,x^3+a}}{6\,x^6}+\frac {b\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )\,\left (A\,b+4\,B\,a\right )}{8\,\sqrt {a}} \] Input:
int(((A + B*x^3)*(a + b*x^3)^(3/2))/x^7,x)
Output:
(2*B*b*(a + b*x^3)^(1/2))/3 - ((a + b*x^3)^(1/2)*(4*B*a^3 + 5*A*a^2*b))/(1 2*a^2*x^3) - (A*a*(a + b*x^3)^(1/2))/(6*x^6) + (b*log((((a + b*x^3)^(1/2) - a^(1/2))^3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6)*(A*b + 4*B*a))/(8*a^(1/2) )
Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx=\frac {-4 \sqrt {b \,x^{3}+a}\, a^{2}-18 \sqrt {b \,x^{3}+a}\, a b \,x^{3}+16 \sqrt {b \,x^{3}+a}\, b^{2} x^{6}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {a}\right ) b^{2} x^{6}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {a}\right ) b^{2} x^{6}}{24 x^{6}} \] Input:
int((b*x^3+a)^(3/2)*(B*x^3+A)/x^7,x)
Output:
( - 4*sqrt(a + b*x**3)*a**2 - 18*sqrt(a + b*x**3)*a*b*x**3 + 16*sqrt(a + b *x**3)*b**2*x**6 + 15*sqrt(a)*log(sqrt(a + b*x**3) - sqrt(a))*b**2*x**6 - 15*sqrt(a)*log(sqrt(a + b*x**3) + sqrt(a))*b**2*x**6)/(24*x**6)