3.1097 \(\int \frac{c+d x^2}{(e x)^{11/2} (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{8 \left (a+b x^2\right )^{5/4} (8 b c-9 a d)}{45 a^3 e^3 (e x)^{5/2}}+\frac{2 \sqrt [4]{a+b x^2} (8 b c-9 a d)}{9 a^2 e^3 (e x)^{5/2}}-\frac{2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}} \]

[Out]

(-2*c*(a + b*x^2)^(1/4))/(9*a*e*(e*x)^(9/2)) + (2*(8*b*c - 9*a*d)*(a + b*x^2)^(1/4))/(9*a^2*e^3*(e*x)^(5/2)) -
(8*(8*b*c - 9*a*d)*(a + b*x^2)^(5/4))/(45*a^3*e^3*(e*x)^(5/2))

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Rubi [A]  time = 0.0508128, 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} \[ -\frac{8 \left (a+b x^2\right )^{5/4} (8 b c-9 a d)}{45 a^3 e^3 (e x)^{5/2}}+\frac{2 \sqrt [4]{a+b x^2} (8 b c-9 a d)}{9 a^2 e^3 (e x)^{5/2}}-\frac{2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*c*(a + b*x^2)^(1/4))/(9*a*e*(e*x)^(9/2)) + (2*(8*b*c - 9*a*d)*(a + b*x^2)^(1/4))/(9*a^2*e^3*(e*x)^(5/2)) -
(8*(8*b*c - 9*a*d)*(a + b*x^2)^(5/4))/(45*a^3*e^3*(e*x)^(5/2))

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)^{11/2} \left (a+b x^2\right )^{3/4}} \, dx &=-\frac{2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}-\frac{(8 b c-9 a d) \int \frac{1}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx}{9 a e^2}\\ &=-\frac{2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}+\frac{2 (8 b c-9 a d) \sqrt [4]{a+b x^2}}{9 a^2 e^3 (e x)^{5/2}}+\frac{(4 (8 b c-9 a d)) \int \frac{\sqrt [4]{a+b x^2}}{(e x)^{7/2}} \, dx}{9 a^2 e^2}\\ &=-\frac{2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}+\frac{2 (8 b c-9 a d) \sqrt [4]{a+b x^2}}{9 a^2 e^3 (e x)^{5/2}}-\frac{8 (8 b c-9 a d) \left (a+b x^2\right )^{5/4}}{45 a^3 e^3 (e x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0397203, size = 72, normalized size = 0.69 \[ -\frac{2 \sqrt{e x} \sqrt [4]{a+b x^2} \left (a^2 \left (5 c+9 d x^2\right )-4 a b x^2 \left (2 c+9 d x^2\right )+32 b^2 c x^4\right )}{45 a^3 e^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*Sqrt[e*x]*(a + b*x^2)^(1/4)*(32*b^2*c*x^4 - 4*a*b*x^2*(2*c + 9*d*x^2) + a^2*(5*c + 9*d*x^2)))/(45*a^3*e^6*
x^5)

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Maple [A]  time = 0.005, size = 62, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( -36\,abd{x}^{4}+32\,{b}^{2}c{x}^{4}+9\,{a}^{2}d{x}^{2}-8\,abc{x}^{2}+5\,{a}^{2}c \right ) }{45\,{a}^{3}}\sqrt [4]{b{x}^{2}+a} \left ( ex \right ) ^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x)

[Out]

-2/45*x*(b*x^2+a)^(1/4)*(-36*a*b*d*x^4+32*b^2*c*x^4+9*a^2*d*x^2-8*a*b*c*x^2+5*a^2*c)/a^3/(e*x)^(11/2)

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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{3}{4}} \left (e x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(11/2)), x)

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Fricas [A]  time = 2.63642, size = 153, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (4 \,{\left (8 \, b^{2} c - 9 \, a b d\right )} x^{4} + 5 \, a^{2} c -{\left (8 \, a b c - 9 \, a^{2} d\right )} x^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{e x}}{45 \, a^{3} e^{6} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

-2/45*(4*(8*b^2*c - 9*a*b*d)*x^4 + 5*a^2*c - (8*a*b*c - 9*a^2*d)*x^2)*(b*x^2 + a)^(1/4)*sqrt(e*x)/(a^3*e^6*x^5
)

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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)**(11/2)/(b*x**2+a)**(3/4),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{3}{4}} \left (e x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(11/2)), x)