3.11.94 \(\int \frac {\sqrt {e x} (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\) [1094]

Optimal. Leaf size=136 \[ \frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e}-\frac {(4 b c-3 a d) \sqrt {e} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac {(4 b c-3 a d) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}} \]

[Out]

1/2*d*(e*x)^(3/2)*(b*x^2+a)^(1/4)/b/e-1/4*(-3*a*d+4*b*c)*arctan(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))*e
^(1/2)/b^(7/4)+1/4*(-3*a*d+4*b*c)*arctanh(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))*e^(1/2)/b^(7/4)

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Rubi [A]
time = 0.06, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {470, 335, 338, 304, 211, 214} \begin {gather*} -\frac {\sqrt {e} (4 b c-3 a d) \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac {\sqrt {e} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(3/4),x]

[Out]

(d*(e*x)^(3/2)*(a + b*x^2)^(1/4))/(2*b*e) - ((4*b*c - 3*a*d)*Sqrt[e]*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a +
b*x^2)^(1/4))])/(4*b^(7/4)) + ((4*b*c - 3*a*d)*Sqrt[e]*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))
])/(4*b^(7/4))

Rule 211

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 470

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

Rubi steps

\begin {align*} \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e}-\frac {\left (-2 b c+\frac {3 a d}{2}\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{3/4}} \, dx}{2 b}\\ &=\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e}+\frac {(4 b c-3 a d) \text {Subst}\left (\int \frac {x^2}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{2 b e}\\ &=\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e}+\frac {(4 b c-3 a d) \text {Subst}\left (\int \frac {x^2}{1-\frac {b x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{2 b e}\\ &=\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e}+\frac {((4 b c-3 a d) e) \text {Subst}\left (\int \frac {1}{e-\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{4 b^{3/2}}-\frac {((4 b c-3 a d) e) \text {Subst}\left (\int \frac {1}{e+\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{4 b^{3/2}}\\ &=\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e}-\frac {(4 b c-3 a d) \sqrt {e} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac {(4 b c-3 a d) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 112, normalized size = 0.82 \begin {gather*} \frac {\sqrt {e x} \left (2 b^{3/4} d x^{3/2} \sqrt [4]{a+b x^2}+(-4 b c+3 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+(4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{7/4} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(3/4),x]

[Out]

(Sqrt[e*x]*(2*b^(3/4)*d*x^(3/2)*(a + b*x^2)^(1/4) + (-4*b*c + 3*a*d)*ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4
)] + (4*b*c - 3*a*d)*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(4*b^(7/4)*Sqrt[x])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {e x}\, \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (87) = 174\).
time = 0.49, size = 189, normalized size = 1.39 \begin {gather*} \frac {1}{8} \, {\left (4 \, c {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} \sqrt {x}}\right )}{b^{\frac {3}{4}}} - \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {x}}}{b^{\frac {1}{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {x}}}\right )}{b^{\frac {3}{4}}}\right )} - d {\left (\frac {3 \, {\left (\frac {2 \, a \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} \sqrt {x}}\right )}{b^{\frac {3}{4}}} - \frac {a \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {x}}}{b^{\frac {1}{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {x}}}\right )}{b^{\frac {3}{4}}}\right )}}{b} + \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} a}{{\left (b^{2} - \frac {{\left (b x^{2} + a\right )} b}{x^{2}}\right )} \sqrt {x}}\right )}\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [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((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C] Result contains complex when optimal does not.
time = 2.07, size = 92, normalized size = 0.68 \begin {gather*} \frac {c \left (e x\right )^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} e \Gamma \left (\frac {7}{4}\right )} + \frac {d \left (e x\right )^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} e^{3} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(1/2)*(d*x**2+c)/(b*x**2+a)**(3/4),x)

[Out]

c*(e*x)**(3/2)*gamma(3/4)*hyper((3/4, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*e*gamma(7/4)) + d*(e
*x)**(7/2)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*e**3*gamma(11/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*sqrt(x)*e^(1/2)/(b*x^2 + a)^(3/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {e\,x}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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