Integrand size = 20, antiderivative size = 94 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\frac {5}{8} a d \sqrt {c+\frac {d}{x^2}} x^2+\frac {1}{4} a c \sqrt {c+\frac {d}{x^2}} x^4+\frac {b \left (c+\frac {d}{x^2}\right )^{5/2} x^5}{5 c}+\frac {3 a d^2 \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 \sqrt {c}} \] Output:
5/8*a*d*(c+d/x^2)^(1/2)*x^2+1/4*a*c*(c+d/x^2)^(1/2)*x^4+1/5*b*(c+d/x^2)^(5 /2)*x^5/c+3/8*a*d^2*arctanh((c+d/x^2)^(1/2)/c^(1/2))/c^(1/2)
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (\sqrt {d+c x^2} \left (8 b \left (d+c x^2\right )^2+5 a c x \left (5 d+2 c x^2\right )\right )-15 a \sqrt {c} d^2 \log \left (-\sqrt {c} x+\sqrt {d+c x^2}\right )\right )}{40 c \sqrt {d+c x^2}} \] Input:
Integrate[(c + d/x^2)^(3/2)*x^3*(a + b*x),x]
Output:
(Sqrt[c + d/x^2]*x*(Sqrt[d + c*x^2]*(8*b*(d + c*x^2)^2 + 5*a*c*x*(5*d + 2* c*x^2)) - 15*a*Sqrt[c]*d^2*Log[-(Sqrt[c]*x) + Sqrt[d + c*x^2]]))/(40*c*Sqr t[d + c*x^2])
Time = 0.44 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1892, 1803, 534, 243, 51, 51, 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.
\(\displaystyle \int x^3 (a+b x) \left (c+\frac {d}{x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1892 |
\(\displaystyle \int x^4 \left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x^2}\right )^{3/2}dx\) |
\(\Big \downarrow \) 1803 |
\(\displaystyle -\int \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {a}{x}+b\right ) x^6d\frac {1}{x}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-a \int \left (c+\frac {d}{x^2}\right )^{3/2} x^5d\frac {1}{x}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-\frac {1}{2} a \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3d\frac {1}{x^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-\frac {1}{2} a \left (\frac {3}{4} d \int \sqrt {c+\frac {d}{x^2}} x^2d\frac {1}{x^2}-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-\frac {1}{2} a \left (\frac {3}{4} d \left (\frac {1}{2} d \int \frac {x}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-\frac {1}{2} a \left (\frac {3}{4} d \left (\int \frac {1}{\frac {\sqrt {c+\frac {d}{x^2}}}{d}-\frac {c}{d}}d\sqrt {c+\frac {d}{x^2}}-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-\frac {1}{2} a \left (\frac {3}{4} d \left (x \left (-\sqrt {c+\frac {d}{x^2}}\right )-\frac {d \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{\sqrt {c}}\right )-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
Input:
Int[(c + d/x^2)^(3/2)*x^3*(a + b*x),x]
Output:
(b*(c + d/x^2)^(5/2)*x^5)/(5*c) - (a*(-1/2*((c + d/x^2)^(3/2)*x^2) + (3*d* (-(Sqrt[c + d/x^2]*x) - (d*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/Sqrt[c]))/4)) /2
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] :> 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_)^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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x )^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^ (p_.), x_Symbol] :> Int[x^(m + mn*q)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; F reeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n 2] || !IntegerQ[p])
Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10
method | result | size |
default | \(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (8 b \left (c \,x^{2}+d \right )^{\frac {5}{2}} \sqrt {c}+10 c^{\frac {3}{2}} \left (c \,x^{2}+d \right )^{\frac {3}{2}} a x +15 c^{\frac {3}{2}} \sqrt {c \,x^{2}+d}\, a d x +15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) a c \,d^{2}\right )}{40 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c^{\frac {3}{2}}}\) | \(103\) |
risch | \(\frac {\left (8 b \,c^{2} x^{4}+10 a \,c^{2} x^{3}+16 b c d \,x^{2}+25 a d x c +8 b \,d^{2}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{40 c}+\frac {3 a \,d^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{8 \sqrt {c}\, \sqrt {c \,x^{2}+d}}\) | \(108\) |
Input:
int((c+d/x^2)^(3/2)*x^3*(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/40*((c*x^2+d)/x^2)^(3/2)*x^3*(8*b*(c*x^2+d)^(5/2)*c^(1/2)+10*c^(3/2)*(c* x^2+d)^(3/2)*a*x+15*c^(3/2)*(c*x^2+d)^(1/2)*a*d*x+15*ln(c^(1/2)*x+(c*x^2+d )^(1/2))*a*c*d^2)/(c*x^2+d)^(3/2)/c^(3/2)
Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.24 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\left [\frac {15 \, a \sqrt {c} d^{2} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) + 2 \, {\left (8 \, b c^{2} x^{5} + 10 \, a c^{2} x^{4} + 16 \, b c d x^{3} + 25 \, a c d x^{2} + 8 \, b d^{2} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{80 \, c}, -\frac {15 \, a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (8 \, b c^{2} x^{5} + 10 \, a c^{2} x^{4} + 16 \, b c d x^{3} + 25 \, a c d x^{2} + 8 \, b d^{2} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{40 \, c}\right ] \] Input:
integrate((c+d/x^2)^(3/2)*x^3*(b*x+a),x, algorithm="fricas")
Output:
[1/80*(15*a*sqrt(c)*d^2*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) + 2*(8*b*c^2*x^5 + 10*a*c^2*x^4 + 16*b*c*d*x^3 + 25*a*c*d*x^2 + 8*b* d^2*x)*sqrt((c*x^2 + d)/x^2))/c, -1/40*(15*a*sqrt(-c)*d^2*arctan(sqrt(-c)* x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (8*b*c^2*x^5 + 10*a*c^2*x^4 + 16* b*c*d*x^3 + 25*a*c*d*x^2 + 8*b*d^2*x)*sqrt((c*x^2 + d)/x^2))/c]
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (87) = 174\).
Time = 3.91 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.10 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\frac {a c^{2} x^{5}}{4 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 a c \sqrt {d} x^{3}}{8 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {a d^{\frac {3}{2}} x \sqrt {\frac {c x^{2}}{d} + 1}}{2} + \frac {a d^{\frac {3}{2}} x}{8 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 a d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 \sqrt {c}} + \frac {b c \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {2 b d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {b d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{5 c} \] Input:
integrate((c+d/x**2)**(3/2)*x**3*(b*x+a),x)
Output:
a*c**2*x**5/(4*sqrt(d)*sqrt(c*x**2/d + 1)) + 3*a*c*sqrt(d)*x**3/(8*sqrt(c* x**2/d + 1)) + a*d**(3/2)*x*sqrt(c*x**2/d + 1)/2 + a*d**(3/2)*x/(8*sqrt(c* x**2/d + 1)) + 3*a*d**2*asinh(sqrt(c)*x/sqrt(d))/(8*sqrt(c)) + b*c*sqrt(d) *x**4*sqrt(c*x**2/d + 1)/5 + 2*b*d**(3/2)*x**2*sqrt(c*x**2/d + 1)/5 + b*d* *(5/2)*sqrt(c*x**2/d + 1)/(5*c)
Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.28 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5}}{5 \, c} - \frac {1}{16} \, {\left (\frac {3 \, d^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{\sqrt {c}} - \frac {2 \, {\left (5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} c + c^{2}}\right )} a \] Input:
integrate((c+d/x^2)^(3/2)*x^3*(b*x+a),x, algorithm="maxima")
Output:
1/5*b*(c + d/x^2)^(5/2)*x^5/c - 1/16*(3*d^2*log((sqrt(c + d/x^2) - sqrt(c) )/(sqrt(c + d/x^2) + sqrt(c)))/sqrt(c) - 2*(5*(c + d/x^2)^(3/2)*d^2 - 3*sq rt(c + d/x^2)*c*d^2)/((c + d/x^2)^2 - 2*(c + d/x^2)*c + c^2))*a
Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.22 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=-\frac {3 \, a d^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + d} \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {c}} + \frac {1}{40} \, \sqrt {c x^{2} + d} {\left (\frac {8 \, b d^{2} \mathrm {sgn}\left (x\right )}{c} + {\left (25 \, a d \mathrm {sgn}\left (x\right ) + 2 \, {\left (8 \, b d \mathrm {sgn}\left (x\right ) + {\left (4 \, b c x \mathrm {sgn}\left (x\right ) + 5 \, a c \mathrm {sgn}\left (x\right )\right )} x\right )} x\right )} x\right )} + \frac {{\left (15 \, a c d^{2} \log \left ({\left | d \right |}\right ) - 16 \, b \sqrt {c} d^{\frac {5}{2}}\right )} \mathrm {sgn}\left (x\right )}{80 \, c^{\frac {3}{2}}} \] Input:
integrate((c+d/x^2)^(3/2)*x^3*(b*x+a),x, algorithm="giac")
Output:
-3/8*a*d^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + d)))*sgn(x)/sqrt(c) + 1/40*sq rt(c*x^2 + d)*(8*b*d^2*sgn(x)/c + (25*a*d*sgn(x) + 2*(8*b*d*sgn(x) + (4*b* c*x*sgn(x) + 5*a*c*sgn(x))*x)*x)*x) + 1/80*(15*a*c*d^2*log(abs(d)) - 16*b* sqrt(c)*d^(5/2))*sgn(x)/c^(3/2)
Time = 7.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {b\,c\,x^5}{5}+\frac {2\,b\,d\,x^3}{5}+\frac {b\,d^2\,x}{5\,c}\right )+\frac {5\,a\,x^4\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{8}+\frac {3\,a\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,\sqrt {c}}-\frac {3\,a\,c\,x^4\,\sqrt {c+\frac {d}{x^2}}}{8} \] Input:
int(x^3*(c + d/x^2)^(3/2)*(a + b*x),x)
Output:
(c + d/x^2)^(1/2)*((b*c*x^5)/5 + (2*b*d*x^3)/5 + (b*d^2*x)/(5*c)) + (5*a*x ^4*(c + d/x^2)^(3/2))/8 + (3*a*d^2*atanh((c + d/x^2)^(1/2)/c^(1/2)))/(8*c^ (1/2)) - (3*a*c*x^4*(c + d/x^2)^(1/2))/8
Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3 (a+b x) \, dx=\frac {10 \sqrt {c \,x^{2}+d}\, a \,c^{2} x^{3}+25 \sqrt {c \,x^{2}+d}\, a c d x +8 \sqrt {c \,x^{2}+d}\, b \,c^{2} x^{4}+16 \sqrt {c \,x^{2}+d}\, b c d \,x^{2}+8 \sqrt {c \,x^{2}+d}\, b \,d^{2}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) a \,d^{2}}{40 c} \] Input:
int((c+d/x^2)^(3/2)*x^3*(b*x+a),x)
Output:
(10*sqrt(c*x**2 + d)*a*c**2*x**3 + 25*sqrt(c*x**2 + d)*a*c*d*x + 8*sqrt(c* x**2 + d)*b*c**2*x**4 + 16*sqrt(c*x**2 + d)*b*c*d*x**2 + 8*sqrt(c*x**2 + d )*b*d**2 + 15*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*a*d**2)/ (40*c)