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

Optimal. Leaf size=67 \[ -\frac {2 \sqrt {e x} (4 b c-3 a d)}{3 a^2 e^3 \sqrt [4]{a+b x^2}}-\frac {2 c}{3 a e (e x)^{3/2} \sqrt [4]{a+b x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {453, 264} \begin {gather*} -\frac {2 \sqrt {e x} (4 b c-3 a d)}{3 a^2 e^3 \sqrt [4]{a+b x^2}}-\frac {2 c}{3 a e (e x)^{3/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 45, normalized size = 0.67 \begin {gather*} \frac {x \left (-2 a c+6 a d x^2-8 b c x^2\right )}{3 a^2 (e x)^{5/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.83, size = 72, normalized size = 1.07 \begin {gather*} \frac {2 \left (a+b x^2\right )^{3/4} \left (-a c e^2+3 a d e^2 x^2-4 b c e^2 x^2\right )}{3 a^2 e (e x)^{3/2} \left (a e^2+b e^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x^2)^(3/4)*(-(a*c*e^2) - 4*b*c*e^2*x^2 + 3*a*d*e^2*x^2))/(3*a^2*e*(e*x)^(3/2)*(a*e^2 + b*e^2*x^2))

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fricas [A]  time = 0.82, size = 57, normalized size = 0.85 \begin {gather*} -\frac {2 \, {\left ({\left (4 \, b c - 3 \, a d\right )} x^{2} + a c\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{3 \, {\left (a^{2} b e^{3} x^{4} + a^{3} 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)^(5/4),x, algorithm="fricas")

[Out]

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

[Out]

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

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maple [A]  time = 0.01, size = 39, normalized size = 0.58 \begin {gather*} -\frac {2 \left (-3 a d \,x^{2}+4 b c \,x^{2}+a c \right ) x}{3 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (e x \right )^{\frac {5}{2}} a^{2}} \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)^(5/4),x)

[Out]

-2/3*x*(-3*a*d*x^2+4*b*c*x^2+a*c)/(b*x^2+a)^(1/4)/a^2/(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 {5}{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)^(5/4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.20, size = 70, normalized size = 1.04 \begin {gather*} -\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,c}{3\,a\,b\,e^2}-\frac {x^2\,\left (6\,a\,d-8\,b\,c\right )}{3\,a^2\,b\,e^2}\right )}{x^3\,\sqrt {e\,x}+\frac {a\,x\,\sqrt {e\,x}}{b}} \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)^(5/4)),x)

[Out]

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

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sympy [A]  time = 64.29, size = 117, normalized size = 1.75 \begin {gather*} c \left (\frac {\Gamma \left (- \frac {3}{4}\right )}{8 a \sqrt [4]{b} e^{\frac {5}{2}} x^{2} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {5}{4}\right )} + \frac {b^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{2 a^{2} e^{\frac {5}{2}} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {5}{4}\right )}\right ) + \frac {d \Gamma \left (\frac {1}{4}\right )}{2 a \sqrt [4]{b} e^{\frac {5}{2}} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {5}{4}\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)**(5/4),x)

[Out]

c*(gamma(-3/4)/(8*a*b**(1/4)*e**(5/2)*x**2*(a/(b*x**2) + 1)**(1/4)*gamma(5/4)) + b**(3/4)*gamma(-3/4)/(2*a**2*
e**(5/2)*(a/(b*x**2) + 1)**(1/4)*gamma(5/4))) + d*gamma(1/4)/(2*a*b**(1/4)*e**(5/2)*(a/(b*x**2) + 1)**(1/4)*ga
mma(5/4))

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