3.1120 \(\int \frac{c+d x^2}{(e x)^{11/2} (a+b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=141 \[ -\frac{64 \left (a+b x^2\right )^{5/4} (4 b c-3 a d)}{45 a^4 e^3 (e x)^{5/2}}+\frac{16 \sqrt [4]{a+b x^2} (4 b c-3 a d)}{9 a^3 e^3 (e x)^{5/2}}-\frac{2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac{2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

(-2*c)/(9*a*e*(e*x)^(9/2)*(a + b*x^2)^(3/4)) - (2*(4*b*c - 3*a*d))/(9*a^2*e^3*(e*x)^(5/2)*(a + b*x^2)^(3/4)) +
(16*(4*b*c - 3*a*d)*(a + b*x^2)^(1/4))/(9*a^3*e^3*(e*x)^(5/2)) - (64*(4*b*c - 3*a*d)*(a + b*x^2)^(5/4))/(45*a
^4*e^3*(e*x)^(5/2))

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Rubi [A]  time = 0.0656395, 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} \[ -\frac{64 \left (a+b x^2\right )^{5/4} (4 b c-3 a d)}{45 a^4 e^3 (e x)^{5/2}}+\frac{16 \sqrt [4]{a+b x^2} (4 b c-3 a d)}{9 a^3 e^3 (e x)^{5/2}}-\frac{2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac{2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(7/4)),x]

[Out]

(-2*c)/(9*a*e*(e*x)^(9/2)*(a + b*x^2)^(3/4)) - (2*(4*b*c - 3*a*d))/(9*a^2*e^3*(e*x)^(5/2)*(a + b*x^2)^(3/4)) +
(16*(4*b*c - 3*a*d)*(a + b*x^2)^(1/4))/(9*a^3*e^3*(e*x)^(5/2)) - (64*(4*b*c - 3*a*d)*(a + b*x^2)^(5/4))/(45*a
^4*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 )^{7/4}} \, dx &=-\frac{2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac{(4 b c-3 a d) \int \frac{1}{(e x)^{7/2} \left (a+b x^2\right )^{7/4}} \, dx}{3 a e^2}\\ &=-\frac{2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac{2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac{(8 (4 b c-3 a d)) \int \frac{1}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx}{9 a^2 e^2}\\ &=-\frac{2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac{2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}+\frac{16 (4 b c-3 a d) \sqrt [4]{a+b x^2}}{9 a^3 e^3 (e x)^{5/2}}+\frac{(32 (4 b c-3 a d)) \int \frac{\sqrt [4]{a+b x^2}}{(e x)^{7/2}} \, dx}{9 a^3 e^2}\\ &=-\frac{2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac{2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}+\frac{16 (4 b c-3 a d) \sqrt [4]{a+b x^2}}{9 a^3 e^3 (e x)^{5/2}}-\frac{64 (4 b c-3 a d) \left (a+b x^2\right )^{5/4}}{45 a^4 e^3 (e x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0401438, size = 88, normalized size = 0.62 \[ \frac{4 x^3 \left (3 a^2-24 a b x^2-32 b^2 x^4\right ) \left (6 b c-\frac{9 a d}{2}\right )}{135 a^4 (e x)^{11/2} \left (a+b x^2\right )^{3/4}}-\frac{2 c x}{9 a (e x)^{11/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(7/4)),x]

[Out]

(-2*c*x)/(9*a*(e*x)^(11/2)*(a + b*x^2)^(3/4)) + (4*(6*b*c - (9*a*d)/2)*x^3*(3*a^2 - 24*a*b*x^2 - 32*b^2*x^4))/
(135*a^4*(e*x)^(11/2)*(a + b*x^2)^(3/4))

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Maple [A]  time = 0.005, size = 86, normalized size = 0.6 \begin{align*} -{\frac{2\,x \left ( -96\,a{b}^{2}d{x}^{6}+128\,{b}^{3}c{x}^{6}-72\,{a}^{2}bd{x}^{4}+96\,a{b}^{2}c{x}^{4}+9\,{a}^{3}d{x}^{2}-12\,{a}^{2}bc{x}^{2}+5\,c{a}^{3} \right ) }{45\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}} \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)^(7/4),x)

[Out]

-2/45*x*(-96*a*b^2*d*x^6+128*b^3*c*x^6-72*a^2*b*d*x^4+96*a*b^2*c*x^4+9*a^3*d*x^2-12*a^2*b*c*x^2+5*a^3*c)/(b*x^
2+a)^(3/4)/a^4/(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{7}{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)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*(e*x)^(11/2)), x)

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Fricas [A]  time = 1.6078, size = 228, normalized size = 1.62 \begin{align*} -\frac{2 \,{\left (32 \,{\left (4 \, b^{3} c - 3 \, a b^{2} d\right )} x^{6} + 24 \,{\left (4 \, a b^{2} c - 3 \, a^{2} b d\right )} x^{4} + 5 \, a^{3} c - 3 \,{\left (4 \, a^{2} b c - 3 \, a^{3} d\right )} x^{2}\right )}{\left (b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{e x}}{45 \,{\left (a^{4} b e^{6} x^{7} + a^{5} e^{6} x^{5}\right )}} \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)^(7/4),x, algorithm="fricas")

[Out]

-2/45*(32*(4*b^3*c - 3*a*b^2*d)*x^6 + 24*(4*a*b^2*c - 3*a^2*b*d)*x^4 + 5*a^3*c - 3*(4*a^2*b*c - 3*a^3*d)*x^2)*
(b*x^2 + a)^(1/4)*sqrt(e*x)/(a^4*b*e^6*x^7 + a^5*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)**(7/4),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{7}{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)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*(e*x)^(11/2)), x)