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

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

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Rubi [A]  time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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]

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 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]

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*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 0.62 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 27.47, size = 100, normalized size = 0.96 \begin {gather*} \frac {2 \left (a+b x^2\right )^{3/4} \left (-5 a^2 c e^4+15 a^2 d e^4 x^2-40 a b c e^4 x^2+12 a b d e^4 x^4-32 b^2 c e^4 x^4\right )}{15 a^3 e (e x)^{3/2} \left (a e^2+b e^2 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.15, size = 95, normalized size = 0.91 \begin {gather*} -\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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 62, normalized size = 0.60 \begin {gather*} -\frac {2 \left (-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 c \,a^{2}\right ) x}{15 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (e x \right )^{\frac {5}{2}} a^{3}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 1.31, size = 115, normalized size = 1.11 \begin {gather*} -\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,c}{3\,a\,b^2\,e^2}-\frac {x^2\,\left (30\,a^2\,d-80\,a\,b\,c\right )}{15\,a^3\,b^2\,e^2}+\frac {x^4\,\left (64\,b^2\,c-24\,a\,b\,d\right )}{15\,a^3\,b^2\,e^2}\right )}{x^5\,\sqrt {e\,x}+\frac {2\,a\,x^3\,\sqrt {e\,x}}{b}+\frac {a^2\,x\,\sqrt {e\,x}}{b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b*x^2)^(3/4)*((2*c)/(3*a*b^2*e^2) - (x^2*(30*a^2*d - 80*a*b*c))/(15*a^3*b^2*e^2) + (x^4*(64*b^2*c - 24*
a*b*d))/(15*a^3*b^2*e^2)))/(x^5*(e*x)^(1/2) + (2*a*x^3*(e*x)^(1/2))/b + (a^2*x*(e*x)^(1/2))/b^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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