3.1110 \(\int \frac{(e x)^{9/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=180 \[ -\frac{7 a^{3/2} e^4 \sqrt{e x} \sqrt [4]{\frac{a}{b x^2}+1} (10 b c-11 a d) E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{20 b^{7/2} \sqrt [4]{a+b x^2}}-\frac{7 a e^3 (e x)^{3/2} (10 b c-11 a d)}{60 b^3 \sqrt [4]{a+b x^2}}+\frac{e (e x)^{7/2} (10 b c-11 a d)}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}} \]

[Out]

(-7*a*(10*b*c - 11*a*d)*e^3*(e*x)^(3/2))/(60*b^3*(a + b*x^2)^(1/4)) + ((10*b*c - 11*a*d)*e*(e*x)^(7/2))/(30*b^
2*(a + b*x^2)^(1/4)) + (d*(e*x)^(11/2))/(5*b*e*(a + b*x^2)^(1/4)) - (7*a^(3/2)*(10*b*c - 11*a*d)*e^4*(1 + a/(b
*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(20*b^(7/2)*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0984225, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {459, 285, 284, 335, 196} \[ -\frac{7 a^{3/2} e^4 \sqrt{e x} \sqrt [4]{\frac{a}{b x^2}+1} (10 b c-11 a d) E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{20 b^{7/2} \sqrt [4]{a+b x^2}}-\frac{7 a e^3 (e x)^{3/2} (10 b c-11 a d)}{60 b^3 \sqrt [4]{a+b x^2}}+\frac{e (e x)^{7/2} (10 b c-11 a d)}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(9/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(-7*a*(10*b*c - 11*a*d)*e^3*(e*x)^(3/2))/(60*b^3*(a + b*x^2)^(1/4)) + ((10*b*c - 11*a*d)*e*(e*x)^(7/2))/(30*b^
2*(a + b*x^2)^(1/4)) + (d*(e*x)^(11/2))/(5*b*e*(a + b*x^2)^(1/4)) - (7*a^(3/2)*(10*b*c - 11*a*d)*e^4*(1 + a/(b
*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(20*b^(7/2)*(a + b*x^2)^(1/4))

Rule 459

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]

Rule 285

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(2*c*(c*x)^(m - 1))/(b*(2*m - 3)*(a + b*x
^2)^(1/4)), x] - Dist[(2*a*c^2*(m - 1))/(b*(2*m - 3)), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

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{(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx &=\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac{\left (-5 b c+\frac{11 a d}{2}\right ) \int \frac{(e x)^{9/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 b}\\ &=\frac{(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac{\left (7 a (10 b c-11 a d) e^2\right ) \int \frac{(e x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{60 b^2}\\ &=-\frac{7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac{(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}+\frac{\left (7 a^2 (10 b c-11 a d) e^4\right ) \int \frac{\sqrt{e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{40 b^3}\\ &=-\frac{7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac{(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}+\frac{\left (7 a^2 (10 b c-11 a d) e^4 \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}{40 b^4 \sqrt [4]{a+b x^2}}\\ &=-\frac{7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac{(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac{\left (7 a^2 (10 b c-11 a d) e^4 \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 )}{40 b^4 \sqrt [4]{a+b x^2}}\\ &=-\frac{7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac{(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac{7 a^{3/2} (10 b c-11 a d) e^4 \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 )}{20 b^{7/2} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.138711, size = 112, normalized size = 0.62 \[ \frac{e^3 (e x)^{3/2} \left (77 a^2 d+7 a \sqrt [4]{\frac{b x^2}{a}+1} (10 b c-11 a d) \, _2F_1\left (\frac{3}{4},\frac{5}{4};\frac{7}{4};-\frac{b x^2}{a}\right )-2 a b \left (35 c+11 d x^2\right )+4 b^2 x^2 \left (5 c+3 d x^2\right )\right )}{60 b^3 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(9/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(e^3*(e*x)^(3/2)*(77*a^2*d + 4*b^2*x^2*(5*c + 3*d*x^2) - 2*a*b*(35*c + 11*d*x^2) + 7*a*(10*b*c - 11*a*d)*(1 +
(b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(60*b^3*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.074, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c) \left ( ex \right ) ^{{\frac{9}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

[Out]

int((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \left (e x\right )^{\frac{9}{2}}}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(9/2)/(b*x^2 + a)^(5/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((d*e^4*x^6 + c*e^4*x^4)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(b^2*x^4 + 2*a*b*x^2 + a^2), 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)**(9/2)*(d*x**2+c)/(b*x**2+a)**(5/4),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 )} \left (e x\right )^{\frac{9}{2}}}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(9/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(9/2)/(b*x^2 + a)^(5/4), x)