3.1130 \(\int \frac{c+d x^2}{(e x)^{9/2} (a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=141 \[ \frac{64 \left (a+b x^2\right )^{3/4} (12 b c-7 a d)}{105 a^4 e^3 (e x)^{3/2}}-\frac{16 (12 b c-7 a d)}{35 a^3 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac{2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}} \]

[Out]

(-2*c)/(7*a*e*(e*x)^(7/2)*(a + b*x^2)^(5/4)) - (2*(12*b*c - 7*a*d))/(35*a^2*e^3*(e*x)^(3/2)*(a + b*x^2)^(5/4))
- (16*(12*b*c - 7*a*d))/(35*a^3*e^3*(e*x)^(3/2)*(a + b*x^2)^(1/4)) + (64*(12*b*c - 7*a*d)*(a + b*x^2)^(3/4))/
(105*a^4*e^3*(e*x)^(3/2))

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Rubi [A]  time = 0.06919, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {453, 273, 264} \[ \frac{64 \left (a+b x^2\right )^{3/4} (12 b c-7 a d)}{105 a^4 e^3 (e x)^{3/2}}-\frac{16 (12 b c-7 a d)}{35 a^3 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac{2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c)/(7*a*e*(e*x)^(7/2)*(a + b*x^2)^(5/4)) - (2*(12*b*c - 7*a*d))/(35*a^2*e^3*(e*x)^(3/2)*(a + b*x^2)^(5/4))
- (16*(12*b*c - 7*a*d))/(35*a^3*e^3*(e*x)^(3/2)*(a + b*x^2)^(1/4)) + (64*(12*b*c - 7*a*d)*(a + b*x^2)^(3/4))/
(105*a^4*e^3*(e*x)^(3/2))

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)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx &=-\frac{2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{(12 b c-7 a d) \int \frac{1}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx}{7 a e^2}\\ &=-\frac{2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{(8 (12 b c-7 a d)) \int \frac{1}{(e x)^{5/2} \left (a+b x^2\right )^{5/4}} \, dx}{35 a^2 e^2}\\ &=-\frac{2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{16 (12 b c-7 a d)}{35 a^3 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac{(32 (12 b c-7 a d)) \int \frac{1}{(e x)^{5/2} \sqrt [4]{a+b x^2}} \, dx}{35 a^3 e^2}\\ &=-\frac{2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac{16 (12 b c-7 a d)}{35 a^3 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}+\frac{64 (12 b c-7 a d) \left (a+b x^2\right )^{3/4}}{105 a^4 e^3 (e x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0483768, size = 94, normalized size = 0.67 \[ \frac{\sqrt{e x} \left (40 a^2 b x^2 \left (3 c-14 d x^2\right )-10 a^3 \left (3 c+7 d x^2\right )+64 a b^2 x^4 \left (15 c-7 d x^2\right )+768 b^3 c x^6\right )}{105 a^4 e^5 x^4 \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*(768*b^3*c*x^6 + 40*a^2*b*x^2*(3*c - 14*d*x^2) + 64*a*b^2*x^4*(15*c - 7*d*x^2) - 10*a^3*(3*c + 7*d*
x^2)))/(105*a^4*e^5*x^4*(a + b*x^2)^(5/4))

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Maple [A]  time = 0.007, size = 86, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( 224\,a{b}^{2}d{x}^{6}-384\,{b}^{3}c{x}^{6}+280\,{a}^{2}bd{x}^{4}-480\,a{b}^{2}c{x}^{4}+35\,{a}^{3}d{x}^{2}-60\,{a}^{2}bc{x}^{2}+15\,c{a}^{3} \right ) }{105\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}} \left ( ex \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*x*(224*a*b^2*d*x^6-384*b^3*c*x^6+280*a^2*b*d*x^4-480*a*b^2*c*x^4+35*a^3*d*x^2-60*a^2*b*c*x^2+15*a^3*c)/
(b*x^2+a)^(5/4)/a^4/(e*x)^(9/2)

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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{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58664, size = 261, normalized size = 1.85 \begin{align*} \frac{2 \,{\left (32 \,{\left (12 \, b^{3} c - 7 \, a b^{2} d\right )} x^{6} + 40 \,{\left (12 \, a b^{2} c - 7 \, a^{2} b d\right )} x^{4} - 15 \, a^{3} c + 5 \,{\left (12 \, a^{2} b c - 7 \, a^{3} d\right )} x^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{e x}}{105 \,{\left (a^{4} b^{2} e^{5} x^{8} + 2 \, a^{5} b e^{5} x^{6} + a^{6} e^{5} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(32*(12*b^3*c - 7*a*b^2*d)*x^6 + 40*(12*a*b^2*c - 7*a^2*b*d)*x^4 - 15*a^3*c + 5*(12*a^2*b*c - 7*a^3*d)*x
^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^4*b^2*e^5*x^8 + 2*a^5*b*e^5*x^6 + a^6*e^5*x^4)

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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)**(9/2)/(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{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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