\(\int \frac {\sqrt {e x} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\) [1112]

Optimal result

Integrand size = 26, antiderivative size = 99 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} b^{3/2} \sqrt [4]{a+b x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {470, 290, 342, 202} \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {\sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-3 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} b^{3/2} \sqrt [4]{a+b x^2}} \]

[In]

Int[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(d*(e*x)^(3/2))/(b*e*(a + b*x^2)^(1/4)) - ((2*b*c - 3*a*d)*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(S
qrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*b^(3/2)*(a + b*x^2)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 290

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[Sqrt[c*x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b
*x^2)^(1/4))), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

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 (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {\left (-b c+\frac {3 a d}{2}\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{b} \\ & = \frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {\left (\left (-b c+\frac {3 a d}{2}\right ) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{b^2 \sqrt [4]{a+b x^2}} \\ & = \frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}+\frac {\left (\left (-b c+\frac {3 a d}{2}\right ) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{b^2 \sqrt [4]{a+b x^2}} \\ & = \frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} b^{3/2} \sqrt [4]{a+b x^2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \sqrt {e x} \left (3 a d+(2 b c-3 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{3 a b \sqrt [4]{a+b x^2}} \]

[In]

Integrate[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(x*Sqrt[e*x]*(3*a*d + (2*b*c - 3*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(
3*a*b*(a + b*x^2)^(1/4))

Maple [F]

\[\int \frac {\sqrt {e x}\, \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

[In]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

[Out]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

Fricas [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.82 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {d \sqrt {e} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} \]

[In]

integrate((e*x)**(1/2)*(d*x**2+c)/(b*x**2+a)**(5/4),x)

[Out]

c*sqrt(e)*x**(3/2)*gamma(3/4)*hyper((3/4, 5/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(7/4)) + d*
sqrt(e)*x**(7/2)*gamma(7/4)*hyper((5/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(11/4))

Maxima [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*sqrt(e*x)/(b*x^2 + a)^(5/4), x)

Giac [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*sqrt(e*x)/(b*x^2 + a)^(5/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {\sqrt {e\,x}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

[In]

int(((e*x)^(1/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x)

[Out]

int(((e*x)^(1/2)*(c + d*x^2))/(a + b*x^2)^(5/4), x)