3.12.17 \(\int \frac {\sqrt {e x} (c+d x^2)}{(a+b x^2)^{7/4}} \, dx\) [1117]

Optimal. Leaf size=125 \[ \frac {2 (b c-a d) (e x)^{3/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {d \sqrt {e} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}} \]

[Out]

2/3*(-a*d+b*c)*(e*x)^(3/2)/a/b/e/(b*x^2+a)^(3/4)-d*arctan(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))*e^(1/2)
/b^(7/4)+d*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 = 0.05, antiderivative size = 125, 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 = {463, 335, 338, 304, 211, 214} \begin {gather*} -\frac {d \sqrt {e} \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac {d \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac {2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(7/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(3/2))/(3*a*b*e*(a + b*x^2)^(3/4)) - (d*Sqrt[e]*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a +
b*x^2)^(1/4))])/b^(7/4) + (d*Sqrt[e]*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/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 463

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

Rubi steps

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

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Mathematica [A]
time = 0.50, size = 108, normalized size = 0.86 \begin {gather*} \frac {\sqrt {e x} \left (\frac {2 b^{3/4} (b c-a d) x^{3/2}}{a \left (a+b x^2\right )^{3/4}}-3 d \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+3 d \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{3 b^{7/4} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(7/4),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {e x}\, \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

[Out]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

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Maxima [A]
time = 0.49, size = 116, normalized size = 0.93 \begin {gather*} \frac {1}{6} \, {\left (d {\left (\frac {3 \, {\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 )}}{b} - \frac {4 \, x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} b}\right )} + \frac {4 \, c x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} a}\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)^(7/4),x, algorithm="maxima")

[Out]

1/6*(d*(3*(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))/b - 4*x^(3/2)/((b*x^2 + a)^(3/4)*b)) + 4*c*x^(3/2)/((b*x^2 + a
)^(3/4)*a))*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)^(7/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C] Result contains complex when optimal does not.
time = 11.01, size = 87, normalized size = 0.70 \begin {gather*} \frac {c \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right )}{2 a^{\frac {7}{4}} \left (1 + \frac {b x^{2}}{a}\right )^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {d \sqrt {e} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \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)**(7/4),x)

[Out]

c*sqrt(e)*x**(3/2)*gamma(3/4)/(2*a**(7/4)*(1 + b*x**2/a)**(3/4)*gamma(7/4)) + d*sqrt(e)*x**(7/2)*gamma(7/4)*hy
per((7/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*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)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*sqrt(x)*e^(1/2)/(b*x^2 + a)^(7/4), 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 )}^{7/4}} \,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)^(7/4),x)

[Out]

int(((e*x)^(1/2)*(c + d*x^2))/(a + b*x^2)^(7/4), x)

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