3.3.78 \(\int x (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\) [278]

Optimal. Leaf size=141 \[ \frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )}{2 d}-\frac {3 \sqrt {b} (b c-a d) e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 d^{5/2}} \]

[Out]

1/2*(e*(b*x^2+a)/(d*x^2+c))^(3/2)*(d*x^2+c)/d-3/2*(-a*d+b*c)*e^(3/2)*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(
1/2)/b^(1/2)/e^(1/2))*b^(1/2)/d^(5/2)+3/2*(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^2

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Rubi [A]
time = 0.10, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1981, 1979, 294, 327, 214} \begin {gather*} -\frac {3 \sqrt {b} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 d^{5/2}}+\frac {3 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac {\left (c+d x^2\right ) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(2*d^2) + (((e*(a + b*x^2))/(c + d*x^2))^(3/2)*(c + d*x^2)
)/(2*d) - (3*Sqrt[b]*(b*c - a*d)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])
])/(2*d^(5/2))

Rule 214

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]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 1979

Int[(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Denominator[p]
}, Dist[q*e*((b*c - a*d)/n), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^(1/n - 1)/(b*e - d*x^q)^(1/n + 1)),
x], x, (e*((a + b*x^n)/(c + d*x^n)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && FractionQ[p] && IntegerQ[1/n
]

Rule 1981

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Dist[1/n, Sub
st[Int[x^(Simplify[(m + 1)/n] - 1)*(e*((a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n,
p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx &=((b c-a d) e) \text {Subst}\left (\int \frac {x^4}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )}{2 d}-\frac {(3 (b c-a d) e) \text {Subst}\left (\int \frac {x^2}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 d}\\ &=\frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )}{2 d}-\frac {\left (3 b (b c-a d) e^2\right ) \text {Subst}\left (\int \frac {1}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 d^2}\\ &=\frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )}{2 d}-\frac {3 \sqrt {b} (b c-a d) e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 d^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 1.40, size = 136, normalized size = 0.96 \begin {gather*} \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {d} \sqrt {a+b x^2} \left (3 b c-2 a d+b d x^2\right )-3 \sqrt {b} (b c-a d) \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )}{2 d^{5/2} \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[d]*Sqrt[a + b*x^2]*(3*b*c - 2*a*d + b*d*x^2) - 3*Sqrt[b]*(b*c - a*d
)*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])]))/(2*d^(5/2)*Sqrt[a + b*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(117)=234\).
time = 0.10, size = 432, normalized size = 3.06

method	result	size
default	\(\frac {\left (3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,d^{2} x^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c d \,x^{2}+2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a c b d -3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}+2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c -4 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a d +4 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b c \right ) \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{4 d^{2} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right )}\)	\(432\)
risch	\(\frac {\left (d \,x^{2}+c \right ) b e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{2 d^{2}}+\frac {\left (\frac {3 b \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a}{4 d \sqrt {d e b}}-\frac {3 b^{2} \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c}{4 d^{2} \sqrt {d e b}}-\frac {x^{2} a^{2} b}{\left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 x^{2} a \,b^{2} c}{d \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {x^{2} b^{3} c^{2}}{d^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {a^{3}}{\left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 a^{2} b c}{d \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {a \,b^{2} c^{2}}{d^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{b \,x^{2}+a}\)	\(513\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 177, normalized size = 1.26 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, {\left (b^{2} c - a b d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{b d^{2} - \frac {{\left (b x^{2} + a\right )} d^{3}}{d x^{2} + c}} + \frac {3 \, {\left (b c - a d\right )} b \log \left (\frac {d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {b d}}{d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {b d}}\right )}{\sqrt {b d} d^{2}} + \frac {4 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{d^{2}}\right )} e^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(b^2*c - a*b*d)*sqrt((b*x^2 + a)/(d*x^2 + c))/(b*d^2 - (b*x^2 + a)*d^3/(d*x^2 + c)) + 3*(b*c - a*d)*b*l
og((d*sqrt((b*x^2 + a)/(d*x^2 + c)) - sqrt(b*d))/(d*sqrt((b*x^2 + a)/(d*x^2 + c)) + sqrt(b*d)))/(sqrt(b*d)*d^2
) + 4*(b*c - a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))/d^2)*e^(3/2)

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Fricas [A]
time = 0.45, size = 303, normalized size = 2.15 \begin {gather*} \left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b}{d}} e^{\frac {3}{2}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {b}{d}}\right ) - 4 \, {\left (b d x^{2} + 3 \, b c - 2 \, a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{8 \, d^{2}}, \frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} x^{2} + a b\right )}}\right ) e^{\frac {3}{2}} + 2 \, {\left (b d x^{2} + 3 \, b c - 2 \, a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{4 \, d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(b*c - a*d)*sqrt(b/d)*e^(3/2)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^
2)*x^2 + 4*(2*b*d^3*x^4 + b*c^2*d + a*c*d^2 + (3*b*c*d^2 + a*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(b/d)
) - 4*(b*d*x^2 + 3*b*c - 2*a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/d^2, 1/4*(3*(b*c - a*d)*sqrt(-b/d)*arct
an(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-b/d)/(b^2*x^2 + a*b))*e^(3/2) + 2*(b*d*x^2
+ 3*b*c - 2*a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/d^2]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (113) = 226\).
time = 5.49, size = 247, normalized size = 1.75 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c} b \mathrm {sgn}\left (d x^{2} + c\right )}{d^{2}} + \frac {4 \, {\left (b^{2} c^{2} \mathrm {sgn}\left (d x^{2} + c\right ) - 2 \, a b c d \mathrm {sgn}\left (d x^{2} + c\right ) + a^{2} d^{2} \mathrm {sgn}\left (d x^{2} + c\right )\right )}}{{\left ({\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} d + \sqrt {b d} c\right )} d^{2}} + \frac {3 \, {\left (\sqrt {b d} b^{2} c \mathrm {sgn}\left (d x^{2} + c\right ) - \sqrt {b d} a b d \mathrm {sgn}\left (d x^{2} + c\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {b d} x^{2} - \sqrt {b d x^{4} + b c x^{2} + a d x^{2} + a c}\right )} b d - \sqrt {b d} b c - \sqrt {b d} a d \right |}\right )}{b d^{3}}\right )} e^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c)*b*sgn(d*x^2 + c)/d^2 + 4*(b^2*c^2*sgn(d*x^2 + c) - 2*a*b*c*d*sg
n(d*x^2 + c) + a^2*d^2*sgn(d*x^2 + c))/(((sqrt(b*d)*x^2 - sqrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c))*d + sqrt(b*
d)*c)*d^2) + 3*(sqrt(b*d)*b^2*c*sgn(d*x^2 + c) - sqrt(b*d)*a*b*d*sgn(d*x^2 + c))*log(abs(-2*(sqrt(b*d)*x^2 - s
qrt(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c))*b*d - sqrt(b*d)*b*c - sqrt(b*d)*a*d))/(b*d^3))*e^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((e*(a + b*x^2))/(c + d*x^2))^(3/2),x)

[Out]

int(x*((e*(a + b*x^2))/(c + d*x^2))^(3/2), x)

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