3.1094 \(\int \frac {\sqrt {e x} (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\)

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

[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 = 0.09, 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 = {459, 329, 331, 298, 205, 208} \[ -\frac {\sqrt {e} (4 b c-3 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac {\sqrt {e} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac {d (e x)^{3/2} \sqrt [4]{a+b x^2}}{2 b e} \]

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 205

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

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 298

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 329

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 331

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 459

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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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 = 0.10, size = 112, normalized size = 0.82 \[ \frac {\sqrt {e x} \left (2 b^{3/4} d x^{3/2} \sqrt [4]{a+b x^2}+(3 a d-4 b c) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+(4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{7/4} \sqrt {x}} \]

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

[Out]

Timed out

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

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(e*x)/(b*x^2 + a)^(3/4), x)

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maple [F]  time = 0.06, 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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

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]

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

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

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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sympy [C]  time = 5.83, size = 92, normalized size = 0.68 \[ \frac {c \left (e x\right )^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} e \Gamma \left (\frac {7}{4}\right )} + \frac {d \left (e x\right )^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} e^{3} \Gamma \left (\frac {11}{4}\right )} \]

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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