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

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

[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 = 0.09, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1960, 288, 321, 208} \[ -\frac {3 \sqrt {b} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 d^{5/2}}+\frac {3 e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac {\left (c+d x^2\right ) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 d} \]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 288

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 321

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 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && 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) \operatorname {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) \operatorname {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 ) \operatorname {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 [C]  time = 0.06, size = 96, normalized size = 0.68 \[ \frac {e \left (a+b x^2\right )^2 \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {d \left (b x^2+a\right )}{a d-b c}\right )}{5 b c-5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(a + b*x^2)^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*Hypergeometric2F1[3/2, 5/
2, 7/2, (d*(a + b*x^2))/(-(b*c) + a*d)])/(5*b*c - 5*a*d)

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fricas [A]  time = 0.73, size = 328, normalized size = 2.33 \[ \left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b e}{d}} e \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (b d e x^{2} + {\left (3 \, b c - 2 \, a d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, d^{2}}, \frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {b e}{d}} e \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} e x^{2} + a b e\right )}}\right ) + 2 \, {\left (b d e x^{2} + {\left (3 \, b c - 2 \, a d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, d^{2}}\right ] \]

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*e/d)*e*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d +
a^2*d^2)*e + 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*e/d)*sqrt((b*e*x^2 + a*e)/(d
*x^2 + c))) - 4*(b*d*e*x^2 + (3*b*c - 2*a*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^2, 1/4*(3*(b*c - a*d)*sqr
t(-b*e/d)*e*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*e/d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*e*x^2 + a*b
*e)) + 2*(b*d*e*x^2 + (3*b*c - 2*a*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^2]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x^2*d+c)]Evaluation time: 0.67Unable to divide, perhaps due to rounding error%%%{%%%{2,[0,3,0]%%%},[2,0,0,0]
%%%}+%%%{%%{[%%%{-4,[0,2,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,0,1]%%%}+%%%{%%%{2,[1,2,1]%%%},[0,0,0,2]%%
%} / %%%{%%%{1,[0,2,2]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-2,[0,1,2]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,0,0,1]%%
%}+%%%{%%%{1,[1,1,3]%%%},[0,0,0,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.05, size = 432, normalized size = 3.06 \[ -\frac {\left (-3 a b \,d^{2} x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{2} c d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-3 a b c d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+3 b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b d \,x^{2}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a d -4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b c -2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c \right ) \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{4 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right ) d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*a*b*d^2
+3*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*x^2*b^2*c*d-2*(b*
d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*d*x^2-3*a*b*c*d*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+3*b^2*c^2*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))+4*((d*x^2+c)*(b*x^2+a))^(1/2)*(b*d)^(1/2)*a*d-4*((d*x^2+c)*(b*x^2+a))^(1/2)*(
b*d)^(1/2)*b*c-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c)/d^2*(d*x^2+c)*((b*x^2+a)/(d*x^2+c)*e)^(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 = 2.21, size = 189, normalized size = 1.34 \[ \frac {1}{4} \, {\left (\frac {2 \, {\left (b^{2} c - a b d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} e}{b d^{2} e - \frac {{\left (b x^{2} + a\right )} d^{3} e}{d x^{2} + c}} + \frac {3 \, {\left (b^{2} c - a b d\right )} e \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} d^{2}} + \frac {4 \, {\left (b c - a d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{d^{2}}\right )} e \]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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