3.1135 \(\int \frac{\sqrt{e x} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0515974, antiderivative size = 114, normalized size of antiderivative = 1., 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 = {457, 284, 335, 196} \[ \frac{2 (e x)^{3/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac{2 \sqrt{e x} \sqrt [4]{\frac{a}{b x^2}+1} (3 a d+2 b c) E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 a^{3/2} b^{3/2} \sqrt [4]{a+b x^2}} \]

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 457

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))]))

Rule 284

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 335

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 196

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

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 ) \operatorname{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*}

Mathematica [C]  time = 0.10117, size = 86, normalized size = 0.75 \[ \frac{x \sqrt{e x} \left (\left (a+b x^2\right ) \sqrt [4]{\frac{b x^2}{a}+1} (3 a d+2 b c) \, _2F_1\left (\frac{3}{4},\frac{9}{4};\frac{7}{4};-\frac{b x^2}{a}\right )-3 a^2 d\right )}{3 a^2 b \left (a+b x^2\right )^{5/4}} \]

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 = 0.031, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c)\sqrt{ex} \left ( b{x}^{2}+a \right ) ^{-{\frac{9}{4}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \sqrt{e x}}{{\left (b x^{2} + a\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

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]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}}{\left (d x^{2} + c\right )} \sqrt{e x}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \end{align*}

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(e*x)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \sqrt{e x}}{{\left (b x^{2} + a\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

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(e*x)/(b*x^2 + a)^(9/4), x)