3.1131 \(\int \frac{c+d x^2}{(e x)^{13/2} (a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=178 \[ -\frac{256 \left (a+b x^2\right )^{7/4} (16 b c-11 a d)}{385 a^5 e^3 (e x)^{7/2}}+\frac{64 \left (a+b x^2\right )^{3/4} (16 b c-11 a d)}{55 a^4 e^3 (e x)^{7/2}}-\frac{24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac{2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}} \]

[Out]

(-2*c)/(11*a*e*(e*x)^(11/2)*(a + b*x^2)^(5/4)) - (2*(16*b*c - 11*a*d))/(55*a^2*e^3*(e*x)^(7/2)*(a + b*x^2)^(5/
4)) - (24*(16*b*c - 11*a*d))/(55*a^3*e^3*(e*x)^(7/2)*(a + b*x^2)^(1/4)) + (64*(16*b*c - 11*a*d)*(a + b*x^2)^(3
/4))/(55*a^4*e^3*(e*x)^(7/2)) - (256*(16*b*c - 11*a*d)*(a + b*x^2)^(7/4))/(385*a^5*e^3*(e*x)^(7/2))

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Rubi [A]  time = 0.0903243, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {453, 273, 264} \[ -\frac{256 \left (a+b x^2\right )^{7/4} (16 b c-11 a d)}{385 a^5 e^3 (e x)^{7/2}}+\frac{64 \left (a+b x^2\right )^{3/4} (16 b c-11 a d)}{55 a^4 e^3 (e x)^{7/2}}-\frac{24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac{2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(9/4)),x]

[Out]

(-2*c)/(11*a*e*(e*x)^(11/2)*(a + b*x^2)^(5/4)) - (2*(16*b*c - 11*a*d))/(55*a^2*e^3*(e*x)^(7/2)*(a + b*x^2)^(5/
4)) - (24*(16*b*c - 11*a*d))/(55*a^3*e^3*(e*x)^(7/2)*(a + b*x^2)^(1/4)) + (64*(16*b*c - 11*a*d)*(a + b*x^2)^(3
/4))/(55*a^4*e^3*(e*x)^(7/2)) - (256*(16*b*c - 11*a*d)*(a + b*x^2)^(7/4))/(385*a^5*e^3*(e*x)^(7/2))

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)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx &=-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac{(16 b c-11 a d) \int \frac{1}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx}{11 a e^2}\\ &=-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{(12 (16 b c-11 a d)) \int \frac{1}{(e x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx}{55 a^2 e^2}\\ &=-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac{(96 (16 b c-11 a d)) \int \frac{1}{(e x)^{9/2} \sqrt [4]{a+b x^2}} \, dx}{55 a^3 e^2}\\ &=-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac{64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}+\frac{(128 (16 b c-11 a d)) \int \frac{\left (a+b x^2\right )^{3/4}}{(e x)^{9/2}} \, dx}{55 a^4 e^2}\\ &=-\frac{2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac{2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac{24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac{64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}-\frac{256 (16 b c-11 a d) \left (a+b x^2\right )^{7/4}}{385 a^5 e^3 (e x)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.054169, size = 79, normalized size = 0.44 \[ \frac{2 x \left (a x^2 \left (20 a^2 b x^2-5 a^3+160 a b^2 x^4+128 b^3 x^6\right ) (11 a d-16 b c)-35 a^5 c\right )}{385 a^6 (e x)^{13/2} \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(9/4)),x]

[Out]

(2*x*(-35*a^5*c + a*(-16*b*c + 11*a*d)*x^2*(-5*a^3 + 20*a^2*b*x^2 + 160*a*b^2*x^4 + 128*b^3*x^6)))/(385*a^6*(e
*x)^(13/2)*(a + b*x^2)^(5/4))

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Maple [A]  time = 0.006, size = 110, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( -1408\,a{b}^{3}d{x}^{8}+2048\,{b}^{4}c{x}^{8}-1760\,{a}^{2}{b}^{2}d{x}^{6}+2560\,a{b}^{3}c{x}^{6}-220\,{a}^{3}bd{x}^{4}+320\,{a}^{2}{b}^{2}c{x}^{4}+55\,{a}^{4}d{x}^{2}-80\,{a}^{3}bc{x}^{2}+35\,c{a}^{4} \right ) }{385\,{a}^{5}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}} \left ( ex \right ) ^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x)

[Out]

-2/385*x*(-1408*a*b^3*d*x^8+2048*b^4*c*x^8-1760*a^2*b^2*d*x^6+2560*a*b^3*c*x^6-220*a^3*b*d*x^4+320*a^2*b^2*c*x
^4+55*a^4*d*x^2-80*a^3*b*c*x^2+35*a^4*c)/(b*x^2+a)^(5/4)/a^5/(e*x)^(13/2)

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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{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(13/2)), x)

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Fricas [A]  time = 1.68508, size = 323, normalized size = 1.81 \begin{align*} -\frac{2 \,{\left (128 \,{\left (16 \, b^{4} c - 11 \, a b^{3} d\right )} x^{8} + 160 \,{\left (16 \, a b^{3} c - 11 \, a^{2} b^{2} d\right )} x^{6} + 35 \, a^{4} c + 20 \,{\left (16 \, a^{2} b^{2} c - 11 \, a^{3} b d\right )} x^{4} - 5 \,{\left (16 \, a^{3} b c - 11 \, a^{4} d\right )} x^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{e x}}{385 \,{\left (a^{5} b^{2} e^{7} x^{10} + 2 \, a^{6} b e^{7} x^{8} + a^{7} e^{7} x^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

-2/385*(128*(16*b^4*c - 11*a*b^3*d)*x^8 + 160*(16*a*b^3*c - 11*a^2*b^2*d)*x^6 + 35*a^4*c + 20*(16*a^2*b^2*c -
11*a^3*b*d)*x^4 - 5*(16*a^3*b*c - 11*a^4*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^5*b^2*e^7*x^10 + 2*a^6*b*e^7*x
^8 + a^7*e^7*x^6)

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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)**(13/2)/(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{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(13/2)), x)