3.3.12 \(\int \frac {x (c+d x^2)^2}{a+b x^2} \, dx\) [212]

Optimal. Leaf size=61 \[ \frac {d (b c-a d) x^2}{2 b^2}+\frac {\left (c+d x^2\right )^2}{4 b}+\frac {(b c-a d)^2 \log \left (a+b x^2\right )}{2 b^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 45} \begin {gather*} \frac {(b c-a d)^2 \log \left (a+b x^2\right )}{2 b^3}+\frac {d x^2 (b c-a d)}{2 b^2}+\frac {\left (c+d x^2\right )^2}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x^2)^2)/(a + b*x^2),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 49, normalized size = 0.80 \begin {gather*} \frac {b d x^2 \left (4 b c-2 a d+b d x^2\right )+2 (b c-a d)^2 \log \left (a+b x^2\right )}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 63, normalized size = 1.03

method	result	size
default	\(-\frac {d \left (-\frac {1}{2} b d \,x^{4}+a d \,x^{2}-2 c \,x^{2} b \right )}{2 b^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(63\)
norman	\(\frac {d^{2} x^{4}}{4 b}-\frac {d \left (a d -2 b c \right ) x^{2}}{2 b^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(64\)
risch	\(\frac {d^{2} x^{4}}{4 b}-\frac {a \,d^{2} x^{2}}{2 b^{2}}+\frac {c d \,x^{2}}{b}+\frac {a^{2} d^{2}}{4 b^{3}}-\frac {a c d}{b^{2}}+\frac {c^{2}}{b}+\frac {\ln \left (b \,x^{2}+a \right ) a^{2} d^{2}}{2 b^{3}}-\frac {\ln \left (b \,x^{2}+a \right ) a c d}{b^{2}}+\frac {\ln \left (b \,x^{2}+a \right ) c^{2}}{2 b}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 66, normalized size = 1.08 \begin {gather*} \frac {b d^{2} x^{4} + 2 \, {\left (2 \, b c d - a d^{2}\right )} x^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.27, size = 67, normalized size = 1.10 \begin {gather*} \frac {b^{2} d^{2} x^{4} + 2 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} x^{2} + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.19, size = 49, normalized size = 0.80 \begin {gather*} x^{2} \left (- \frac {a d^{2}}{2 b^{2}} + \frac {c d}{b}\right ) + \frac {d^{2} x^{4}}{4 b} + \frac {\left (a d - b c\right )^{2} \log {\left (a + b x^{2} \right )}}{2 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.98, size = 67, normalized size = 1.10 \begin {gather*} \frac {b d^{2} x^{4} + 4 \, b c d x^{2} - 2 \, a d^{2} x^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 68, normalized size = 1.11 \begin {gather*} \frac {d^2\,x^4}{4\,b}-x^2\,\left (\frac {a\,d^2}{2\,b^2}-\frac {c\,d}{b}\right )+\frac {\ln \left (b\,x^2+a\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(c + d*x^2)^2)/(a + b*x^2),x)

[Out]

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

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