\(\int \frac {c+d x^2}{\sqrt {e x} (a+b x^2)^{3/4}} \, dx\) [1101]

Optimal result

Integrand size = 26, antiderivative size = 102 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx=\frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}-\frac {(2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \sqrt {b} e^2 \left (a+b x^2\right )^{3/4}} \]

[Out]

-(-a*d+2*b*c)*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1
/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))/e^2/(b*x^2+a)^(3/4)/a^(1/2)/b^(1/2)+d*(b*x
^2+a)^(1/4)*(e*x)^(1/2)/b/e

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {470, 335, 243, 342, 281, 237} \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx=\frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}-\frac {(e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-a d) \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \sqrt {b} e^2 \left (a+b x^2\right )^{3/4}} \]

[In]

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

[Out]

(d*Sqrt[e*x]*(a + b*x^2)^(1/4))/(b*e) - ((2*b*c - a*d)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcCot[(Sqr
t[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*Sqrt[b]*e^2*(a + b*x^2)^(3/4))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 281

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

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 342

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

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*} \text {integral}& = \frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}-\frac {\left (-b c+\frac {a d}{2}\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{b} \\ & = \frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}+\frac {(2 b c-a d) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{b e} \\ & = \frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}+\frac {\left ((2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {e x}\right )}{b e \left (a+b x^2\right )^{3/4}} \\ & = \frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}-\frac {\left ((2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a e^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{b e \left (a+b x^2\right )^{3/4}} \\ & = \frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}-\frac {\left ((2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{e x}\right )}{2 b e \left (a+b x^2\right )^{3/4}} \\ & = \frac {d \sqrt {e x} \sqrt [4]{a+b x^2}}{b e}-\frac {(2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} e^2 \left (a+b x^2\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx=\frac {d x \left (a+b x^2\right )+(2 b c-a d) x \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^2}{a}\right )}{b \sqrt {e x} \left (a+b x^2\right )^{3/4}} \]

[In]

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

[Out]

(d*x*(a + b*x^2) + (2*b*c - a*d)*x*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x^2)/a)])/(b*Sq
rt[e*x]*(a + b*x^2)^(3/4))

Maple [F]

\[\int \frac {d \,x^{2}+c}{\sqrt {e x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.41 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{b^{\frac {3}{4}} \sqrt {e} x} + \frac {d x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

-c*hyper((1/2, 3/4), (3/2,), a*exp_polar(I*pi)/(b*x**2))/(b**(3/4)*sqrt(e)*x) + d*x**(5/2)*gamma(5/4)*hyper((3
/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*sqrt(e)*gamma(9/4))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx=\int \frac {d\,x^2+c}{\sqrt {e\,x}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \]

[In]

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

[Out]

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