3.1129 \(\int \frac{c+d x^2}{(e x)^{5/2} (a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{8 \sqrt{e x} (8 b c-3 a d)}{15 a^3 e^3 \sqrt [4]{a+b x^2}}-\frac{2 \sqrt{e x} (8 b c-3 a d)}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac{2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}} \]

[Out]

(-2*c)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(5/4)) - (2*(8*b*c - 3*a*d)*Sqrt[e*x])/(15*a^2*e^3*(a + b*x^2)^(5/4)) -
(8*(8*b*c - 3*a*d)*Sqrt[e*x])/(15*a^3*e^3*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0481076, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {453, 273, 264} \[ -\frac{8 \sqrt{e x} (8 b c-3 a d)}{15 a^3 e^3 \sqrt [4]{a+b x^2}}-\frac{2 \sqrt{e x} (8 b c-3 a d)}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac{2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(5/4)) - (2*(8*b*c - 3*a*d)*Sqrt[e*x])/(15*a^2*e^3*(a + b*x^2)^(5/4)) -
(8*(8*b*c - 3*a*d)*Sqrt[e*x])/(15*a^3*e^3*(a + b*x^2)^(1/4))

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 273

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx &=-\frac{2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{(8 b c-3 a d) \int \frac{1}{\sqrt{e x} \left (a+b x^2\right )^{9/4}} \, dx}{3 a e^2}\\ &=-\frac{2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (8 b c-3 a d) \sqrt{e x}}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac{(4 (8 b c-3 a d)) \int \frac{1}{\sqrt{e x} \left (a+b x^2\right )^{5/4}} \, dx}{15 a^2 e^2}\\ &=-\frac{2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (8 b c-3 a d) \sqrt{e x}}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac{8 (8 b c-3 a d) \sqrt{e x}}{15 a^3 e^3 \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0347551, size = 65, normalized size = 0.62 \[ \frac{x \left (-10 a^2 \left (c-3 d x^2\right )+a b \left (24 d x^4-80 c x^2\right )-64 b^2 c x^4\right )}{15 a^3 (e x)^{5/2} \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-64*b^2*c*x^4 - 10*a^2*(c - 3*d*x^2) + a*b*(-80*c*x^2 + 24*d*x^4)))/(15*a^3*(e*x)^(5/2)*(a + b*x^2)^(5/4))

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Maple [A]  time = 0.004, size = 62, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( -12\,abd{x}^{4}+32\,{b}^{2}c{x}^{4}-15\,{a}^{2}d{x}^{2}+40\,abc{x}^{2}+5\,{a}^{2}c \right ) }{15\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}} \left ( ex \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*x*(-12*a*b*d*x^4+32*b^2*c*x^4-15*a^2*d*x^2+40*a*b*c*x^2+5*a^2*c)/(b*x^2+a)^(5/4)/a^3/(e*x)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82257, size = 204, normalized size = 1.96 \begin{align*} -\frac{2 \,{\left (4 \,{\left (8 \, b^{2} c - 3 \, a b d\right )} x^{4} + 5 \, a^{2} c + 5 \,{\left (8 \, a b c - 3 \, a^{2} d\right )} x^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{e x}}{15 \,{\left (a^{3} b^{2} e^{3} x^{6} + 2 \, a^{4} b e^{3} x^{4} + a^{5} e^{3} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(4*(8*b^2*c - 3*a*b*d)*x^4 + 5*a^2*c + 5*(8*a*b*c - 3*a^2*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^3*b^2*e
^3*x^6 + 2*a^4*b*e^3*x^4 + a^5*e^3*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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