3.1128 \(\int \frac {c+d x^2}{\sqrt {e x} (a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=79 \[ \frac {2 \sqrt {e x} (a d+4 b c)}{5 a^2 b e \sqrt [4]{a+b x^2}}+\frac {2 \sqrt {e x} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 79, 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 = {457, 264} \[ \frac {2 \sqrt {e x} (a d+4 b c)}{5 a^2 b e \sqrt [4]{a+b x^2}}+\frac {2 \sqrt {e x} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c - a*d)*Sqrt[e*x])/(5*a*b*e*(a + b*x^2)^(5/4)) + (2*(4*b*c + a*d)*Sqrt[e*x])/(5*a^2*b*e*(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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 44, normalized size = 0.56 \[ \frac {2 x \left (5 a c+a d x^2+4 b c x^2\right )}{5 a^2 \sqrt {e x} \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 62, normalized size = 0.78 \[ \frac {2 \, {\left ({\left (4 \, b c + a d\right )} x^{2} + 5 \, a c\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{5 \, {\left (a^{2} b^{2} e x^{4} + 2 \, a^{3} b e x^{2} + a^{4} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

2/5*((4*b*c + a*d)*x^2 + 5*a*c)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^2*b^2*e*x^4 + 2*a^3*b*e*x^2 + a^4*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \sqrt {e x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 39, normalized size = 0.49 \[ \frac {2 \left (a d \,x^{2}+4 b c \,x^{2}+5 a c \right ) x}{5 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \sqrt {e x}\, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(9/4),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \sqrt {e x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.21, size = 79, normalized size = 1.00 \[ \frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {x^3\,\left (2\,a\,d+8\,b\,c\right )}{5\,a^2\,b^2}+\frac {2\,c\,x}{a\,b^2}\right )}{x^4\,\sqrt {e\,x}+\frac {a^2\,\sqrt {e\,x}}{b^2}+\frac {2\,a\,x^2\,\sqrt {e\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(1/2)*(a + b*x^2)^(9/4)),x)

[Out]

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

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sympy [B]  time = 165.87, size = 230, normalized size = 2.91 \[ c \left (\frac {5 a \Gamma \left (\frac {1}{4}\right )}{8 a^{3} \sqrt [4]{b} \sqrt {e} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {9}{4}\right ) + 8 a^{2} b^{\frac {5}{4}} \sqrt {e} x^{2} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {9}{4}\right )} + \frac {4 b x^{2} \Gamma \left (\frac {1}{4}\right )}{8 a^{3} \sqrt [4]{b} \sqrt {e} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {9}{4}\right ) + 8 a^{2} b^{\frac {5}{4}} \sqrt {e} x^{2} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {9}{4}\right )}\right ) + \frac {d \Gamma \left (\frac {5}{4}\right )}{\frac {2 a^{2} \sqrt [4]{b} \sqrt {e} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {9}{4}\right )}{x^{2}} + 2 a b^{\frac {5}{4}} \sqrt {e} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(e*x)**(1/2)/(b*x**2+a)**(9/4),x)

[Out]

c*(5*a*gamma(1/4)/(8*a**3*b**(1/4)*sqrt(e)*(a/(b*x**2) + 1)**(1/4)*gamma(9/4) + 8*a**2*b**(5/4)*sqrt(e)*x**2*(
a/(b*x**2) + 1)**(1/4)*gamma(9/4)) + 4*b*x**2*gamma(1/4)/(8*a**3*b**(1/4)*sqrt(e)*(a/(b*x**2) + 1)**(1/4)*gamm
a(9/4) + 8*a**2*b**(5/4)*sqrt(e)*x**2*(a/(b*x**2) + 1)**(1/4)*gamma(9/4))) + d*gamma(5/4)/(2*a**2*b**(1/4)*sqr
t(e)*(a/(b*x**2) + 1)**(1/4)*gamma(9/4)/x**2 + 2*a*b**(5/4)*sqrt(e)*(a/(b*x**2) + 1)**(1/4)*gamma(9/4))

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