3.792 \(\int \frac{\sqrt{c+d x^4}}{x^3 (a+b x^4)} \, dx\)

Optimal. Leaf size=76 \[ -\frac{\sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{3/2}}-\frac{\sqrt{c+d x^4}}{2 a x^2} \]

[Out]

-Sqrt[c + d*x^4]/(2*a*x^2) - (Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(2*a^(3
/2))

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Rubi [A]  time = 0.0862588, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {465, 475, 12, 377, 205} \[ -\frac{\sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{3/2}}-\frac{\sqrt{c+d x^4}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[c + d*x^4]/(2*a*x^2) - (Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(2*a^(3
/2))

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d x^4}}{x^3 \left (a+b x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{c+d x^2}}{x^2 \left (a+b x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{c+d x^4}}{2 a x^2}+\frac{\operatorname{Subst}\left (\int \frac{-b c+a d}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^2\right )}{2 a}\\ &=-\frac{\sqrt{c+d x^4}}{2 a x^2}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^2\right )}{2 a}\\ &=-\frac{\sqrt{c+d x^4}}{2 a x^2}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x^2}{\sqrt{c+d x^4}}\right )}{2 a}\\ &=-\frac{\sqrt{c+d x^4}}{2 a x^2}-\frac{\sqrt{b c-a d} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^2}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0177218, size = 53, normalized size = 0.7 \[ -\frac{\sqrt{c+d x^4} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{(a d-b c) x^4}{a \left (d x^4+c\right )}\right )}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x^4]*Hypergeometric2F1[-1/2, 1, 1/2, ((-(b*c) + a*d)*x^4)/(a*(c + d*x^4))])/(2*a*x^2)

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Maple [B]  time = 0.016, size = 1075, normalized size = 14.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^4+c)^(1/2)/x^3/(b*x^4+a),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{4} + c}}{{\left (b x^{4} + a\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^4 + c)/((b*x^4 + a)*x^3), x)

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Fricas [A]  time = 1.65851, size = 586, normalized size = 7.71 \begin{align*} \left [\frac{x^{2} \sqrt{-\frac{b c - a d}{a}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} - 4 \,{\left ({\left (a b c - 2 \, a^{2} d\right )} x^{6} - a^{2} c x^{2}\right )} \sqrt{d x^{4} + c} \sqrt{-\frac{b c - a d}{a}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) - 4 \, \sqrt{d x^{4} + c}}{8 \, a x^{2}}, -\frac{x^{2} \sqrt{\frac{b c - a d}{a}} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c} \sqrt{\frac{b c - a d}{a}}}{2 \,{\left ({\left (b c d - a d^{2}\right )} x^{6} +{\left (b c^{2} - a c d\right )} x^{2}\right )}}\right ) + 2 \, \sqrt{d x^{4} + c}}{4 \, a x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^3/(b*x^4+a),x, algorithm="fricas")

[Out]

[1/8*(x^2*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4 + a^
2*c^2 - 4*((a*b*c - 2*a^2*d)*x^6 - a^2*c*x^2)*sqrt(d*x^4 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^8 + 2*a*b*x^4 + a^2
)) - 4*sqrt(d*x^4 + c))/(a*x^2), -1/4*(x^2*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4
+ c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^6 + (b*c^2 - a*c*d)*x^2)) + 2*sqrt(d*x^4 + c))/(a*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{4}}}{x^{3} \left (a + b x^{4}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**4+c)**(1/2)/x**3/(b*x**4+a),x)

[Out]

Integral(sqrt(c + d*x**4)/(x**3*(a + b*x**4)), x)

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Giac [A]  time = 1.13476, size = 89, normalized size = 1.17 \begin{align*} \frac{{\left (b c - a d\right )} \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{4}}}}{\sqrt{a b c - a^{2} d}}\right )}{2 \, \sqrt{a b c - a^{2} d} a} - \frac{\sqrt{d + \frac{c}{x^{4}}}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^3/(b*x^4+a),x, algorithm="giac")

[Out]

1/2*(b*c - a*d)*arctan(a*sqrt(d + c/x^4)/sqrt(a*b*c - a^2*d))/(sqrt(a*b*c - a^2*d)*a) - 1/2*sqrt(d + c/x^4)/a