3.12.35 \(\int \frac {\sqrt {e x} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [1135]

Optimal. Leaf size=114 \[ \frac {2 (b c-a d) (e x)^{3/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c+3 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} b^{3/2} \sqrt [4]{a+b x^2}} \]

[Out]

2/5*(-a*d+b*c)*(e*x)^(3/2)/a/b/e/(b*x^2+a)^(5/4)-2/5*(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)/a^(3/2)/b^(3/2)/(b*x^2+a)^(1/4)

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Rubi [A]
time = 0.03, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {468, 290, 342, 202} \begin {gather*} \frac {2 (e x)^{3/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (3 a d+2 b c) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} b^{3/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(3/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - (2*(2*b*c + 3*a*d)*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*E
llipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/2)*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 468

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*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rubi steps

\begin {align*} \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx &=\frac {2 (b c-a d) (e x)^{3/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {\left (2 \left (b c+\frac {3 a d}{2}\right )\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a b}\\ &=\frac {2 (b c-a d) (e x)^{3/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {\left (2 \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}{5 a b^2 \sqrt [4]{a+b x^2}}\\ &=\frac {2 (b c-a d) (e x)^{3/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {\left (2 \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 )}{5 a b^2 \sqrt [4]{a+b x^2}}\\ &=\frac {2 (b c-a d) (e x)^{3/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 (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 )}{5 a^{3/2} b^{3/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.09, size = 86, normalized size = 0.75 \begin {gather*} \frac {x \sqrt {e x} \left (-3 a^2 d+(2 b c+3 a d) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {3}{4},\frac {9}{4};\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{3 a^2 b \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(x*Sqrt[e*x]*(-3*a^2*d + (2*b*c + 3*a*d)*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 9/4, 7/4, -(
(b*x^2)/a)]))/(3*a^2*b*(a + b*x^2)^(5/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {e x}\, \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

[Out]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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)^(9/4),x, algorithm="maxima")

[Out]

e^(1/2)*integrate((d*x^2 + c)*sqrt(x)/(b*x^2 + a)^(9/4), x)

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Fricas [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)^(9/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(x)*e^(1/2)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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Sympy [C] Result contains complex when optimal does not.
time = 44.57, size = 94, normalized size = 0.82 \begin {gather*} \frac {c \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{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 {9}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{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)**(9/4),x)

[Out]

c*sqrt(e)*x**(3/2)*gamma(3/4)*hyper((3/4, 9/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(9/4)*gamma(7/4)) + d*
sqrt(e)*x**(7/2)*gamma(7/4)*hyper((7/4, 9/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(9/4)*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)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*sqrt(x)*e^(1/2)/(b*x^2 + a)^(9/4), 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 )}^{9/4}} \,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)^(9/4),x)

[Out]

int(((e*x)^(1/2)*(c + d*x^2))/(a + b*x^2)^(9/4), x)

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