\(\int \frac {(c+d x^2)^3}{x (a+b x^2)} \, dx\) [224]

Optimal result

Integrand size = 22, antiderivative size = 73 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=\frac {d^2 (3 b c-a d) x^2}{2 b^2}+\frac {d^3 x^4}{4 b}+\frac {c^3 \log (x)}{a}-\frac {(b c-a d)^3 \log \left (a+b x^2\right )}{2 a b^3} \]

[Out]

1/2*d^2*(-a*d+3*b*c)*x^2/b^2+1/4*d^3*x^4/b+c^3*ln(x)/a-1/2*(-a*d+b*c)^3*ln(b*x^2+a)/a/b^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 84} \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=-\frac {(b c-a d)^3 \log \left (a+b x^2\right )}{2 a b^3}+\frac {d^2 x^2 (3 b c-a d)}{2 b^2}+\frac {c^3 \log (x)}{a}+\frac {d^3 x^4}{4 b} \]

[In]

Int[(c + d*x^2)^3/(x*(a + b*x^2)),x]

[Out]

(d^2*(3*b*c - a*d)*x^2)/(2*b^2) + (d^3*x^4)/(4*b) + (c^3*Log[x])/a - ((b*c - a*d)^3*Log[a + b*x^2])/(2*a*b^3)

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^3}{x (a+b x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {d^2 (3 b c-a d)}{b^2}+\frac {c^3}{a x}+\frac {d^3 x}{b}+\frac {(-b c+a d)^3}{a b^2 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {d^2 (3 b c-a d) x^2}{2 b^2}+\frac {d^3 x^4}{4 b}+\frac {c^3 \log (x)}{a}-\frac {(b c-a d)^3 \log \left (a+b x^2\right )}{2 a b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=\frac {a b d^2 x^2 \left (6 b c-2 a d+b d x^2\right )+4 b^3 c^3 \log (x)-2 (b c-a d)^3 \log \left (a+b x^2\right )}{4 a b^3} \]

[In]

Integrate[(c + d*x^2)^3/(x*(a + b*x^2)),x]

[Out]

(a*b*d^2*x^2*(6*b*c - 2*a*d + b*d*x^2) + 4*b^3*c^3*Log[x] - 2*(b*c - a*d)^3*Log[a + b*x^2])/(4*a*b^3)

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {d \left (-b d \,x^{2}+a d -3 b c \right )^{2}}{4 b^{3}}+\frac {c^{3} \ln \left (x \right )}{a}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b \,x^{2}+a \right )}{2 a \,b^{3}}\)	\(86\)
norman	\(\frac {d^{3} x^{4}}{4 b}-\frac {d^{2} \left (a d -3 b c \right ) x^{2}}{2 b^{2}}+\frac {c^{3} \ln \left (x \right )}{a}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (b \,x^{2}+a \right )}{2 a \,b^{3}}\)	\(93\)
parallelrisch	\(\frac {a \,b^{2} d^{3} x^{4}-2 x^{2} a^{2} b \,d^{3}+6 x^{2} a \,b^{2} c \,d^{2}+4 c^{3} \ln \left (x \right ) b^{3}+2 \ln \left (b \,x^{2}+a \right ) a^{3} d^{3}-6 \ln \left (b \,x^{2}+a \right ) a^{2} b c \,d^{2}+6 \ln \left (b \,x^{2}+a \right ) a \,b^{2} c^{2} d -2 \ln \left (b \,x^{2}+a \right ) b^{3} c^{3}}{4 a \,b^{3}}\)	\(124\)
risch	\(\frac {d^{3} x^{4}}{4 b}-\frac {d^{3} x^{2} a}{2 b^{2}}+\frac {3 d^{2} x^{2} c}{2 b}+\frac {d^{3} a^{2}}{4 b^{3}}-\frac {3 d^{2} a c}{2 b^{2}}+\frac {9 d \,c^{2}}{4 b}+\frac {c^{3} \ln \left (x \right )}{a}+\frac {a^{2} \ln \left (-b \,x^{2}-a \right ) d^{3}}{2 b^{3}}-\frac {3 a \ln \left (-b \,x^{2}-a \right ) c \,d^{2}}{2 b^{2}}+\frac {3 \ln \left (-b \,x^{2}-a \right ) c^{2} d}{2 b}-\frac {\ln \left (-b \,x^{2}-a \right ) c^{3}}{2 a}\)	\(158\)

[In]

int((d*x^2+c)^3/x/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

1/4*d*(-b*d*x^2+a*d-3*b*c)^2/b^3+c^3*ln(x)/a+1/2*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/a/b^3*ln(b*x^2+
a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.38 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=\frac {a b^{2} d^{3} x^{4} + 4 \, b^{3} c^{3} \log \left (x\right ) + 2 \, {\left (3 \, a b^{2} c d^{2} - a^{2} b d^{3}\right )} x^{2} - 2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (b x^{2} + a\right )}{4 \, a b^{3}} \]

[In]

integrate((d*x^2+c)^3/x/(b*x^2+a),x, algorithm="fricas")

[Out]

1/4*(a*b^2*d^3*x^4 + 4*b^3*c^3*log(x) + 2*(3*a*b^2*c*d^2 - a^2*b*d^3)*x^2 - 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2
*b*c*d^2 - a^3*d^3)*log(b*x^2 + a))/(a*b^3)

Sympy [A] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=x^{2} \left (- \frac {a d^{3}}{2 b^{2}} + \frac {3 c d^{2}}{2 b}\right ) + \frac {d^{3} x^{4}}{4 b} + \frac {c^{3} \log {\left (x \right )}}{a} + \frac {\left (a d - b c\right )^{3} \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a b^{3}} \]

[In]

integrate((d*x**2+c)**3/x/(b*x**2+a),x)

[Out]

x**2*(-a*d**3/(2*b**2) + 3*c*d**2/(2*b)) + d**3*x**4/(4*b) + c**3*log(x)/a + (a*d - b*c)**3*log(a/b + x**2)/(2
*a*b**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.34 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=\frac {c^{3} \log \left (x^{2}\right )}{2 \, a} + \frac {b d^{3} x^{4} + 2 \, {\left (3 \, b c d^{2} - a d^{3}\right )} x^{2}}{4 \, b^{2}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (b x^{2} + a\right )}{2 \, a b^{3}} \]

[In]

integrate((d*x^2+c)^3/x/(b*x^2+a),x, algorithm="maxima")

[Out]

1/2*c^3*log(x^2)/a + 1/4*(b*d^3*x^4 + 2*(3*b*c*d^2 - a*d^3)*x^2)/b^2 - 1/2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*
c*d^2 - a^3*d^3)*log(b*x^2 + a)/(a*b^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=\frac {c^{3} \log \left (x^{2}\right )}{2 \, a} + \frac {b d^{3} x^{4} + 6 \, b c d^{2} x^{2} - 2 \, a d^{3} x^{2}}{4 \, b^{2}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a b^{3}} \]

[In]

integrate((d*x^2+c)^3/x/(b*x^2+a),x, algorithm="giac")

[Out]

1/2*c^3*log(x^2)/a + 1/4*(b*d^3*x^4 + 6*b*c*d^2*x^2 - 2*a*d^3*x^2)/b^2 - 1/2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*
b*c*d^2 - a^3*d^3)*log(abs(b*x^2 + a))/(a*b^3)

Mupad [B] (verification not implemented)

Time = 5.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \frac {\left (c+d x^2\right )^3}{x \left (a+b x^2\right )} \, dx=\frac {d^3\,x^4}{4\,b}-x^2\,\left (\frac {a\,d^3}{2\,b^2}-\frac {3\,c\,d^2}{2\,b}\right )+\frac {c^3\,\ln \left (x\right )}{a}+\frac {\ln \left (b\,x^2+a\right )\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{2\,a\,b^3} \]

[In]

int((c + d*x^2)^3/(x*(a + b*x^2)),x)

[Out]

(d^3*x^4)/(4*b) - x^2*((a*d^3)/(2*b^2) - (3*c*d^2)/(2*b)) + (c^3*log(x))/a + (log(a + b*x^2)*(a^3*d^3 - b^3*c^
3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(2*a*b^3)