3.1113 \(\int \frac {c+d x^2}{(e x)^{3/2} (a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=103 \[ \frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt {b} e^2 \sqrt [4]{a+b x^2}}-\frac {2 c}{a e \sqrt {e x} \sqrt [4]{a+b x^2}} \]

[Out]

-2*c/a/e/(b*x^2+a)^(1/4)/(e*x)^(1/2)+2*(-a*d+2*b*c)*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(
1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*(e*x)^(1/2)/a^(3
/2)/e^2/(b*x^2+a)^(1/4)/b^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {453, 284, 335, 196} \[ \frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt {b} e^2 \sqrt [4]{a+b x^2}}-\frac {2 c}{a e \sqrt {e x} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(3/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*c)/(a*e*Sqrt[e*x]*(a + b*x^2)^(1/4)) + (2*(2*b*c - a*d)*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(
Sqrt[b]*x)/Sqrt[a]]/2, 2])/(a^(3/2)*Sqrt[b]*e^2*(a + b*x^2)^(1/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 284

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[(Sqrt[c*x]*(1 + a/(b*x^2))^(1/4))/(b*(a +
b*x^2)^(1/4)), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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

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Mathematica [C]  time = 0.04, size = 77, normalized size = 0.75 \[ \frac {x \left (2 x^2 \sqrt [4]{\frac {b x^2}{a}+1} (a d-2 b c) \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};-\frac {b x^2}{a}\right )-6 a c\right )}{3 a^2 (e x)^{3/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(3/2)*(a + b*x^2)^(5/4)),x]

[Out]

(x*(-6*a*c + 2*(-2*b*c + a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(3*a^
2*(e*x)^(3/2)*(a + b*x^2)^(1/4))

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )} \sqrt {e x}}{b^{2} e^{2} x^{6} + 2 \, a b e^{2} x^{4} + a^{2} e^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(e*x)/(b^2*e^2*x^6 + 2*a*b*e^2*x^4 + a^2*e^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {d\,x^2+c}{{\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 38.28, size = 82, normalized size = 0.80 \[ - \frac {d {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{b^{\frac {5}{4}} e^{\frac {3}{2}} x} + \frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(e*x)**(3/2)/(b*x**2+a)**(5/4),x)

[Out]

-d*hyper((1/2, 5/4), (3/2,), a*exp_polar(I*pi)/(b*x**2))/(b**(5/4)*e**(3/2)*x) + c*gamma(-1/4)*hyper((-1/4, 5/
4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*e**(3/2)*sqrt(x)*gamma(3/4))

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