3.794 \(\int \frac{\sqrt{c+d x^4}}{x^7 (a+b x^4)} \, dx\)

Optimal. Leaf size=110 \[ \frac{\sqrt{c+d x^4} (3 b c-a d)}{6 a^2 c x^2}+\frac{b \sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{5/2}}-\frac{\sqrt{c+d x^4}}{6 a x^6} \]

[Out]

-Sqrt[c + d*x^4]/(6*a*x^6) + ((3*b*c - a*d)*Sqrt[c + d*x^4])/(6*a^2*c*x^2) + (b*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b
*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(2*a^(5/2))

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Rubi [A]  time = 0.15976, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {465, 475, 583, 12, 377, 205} \[ \frac{\sqrt{c+d x^4} (3 b c-a d)}{6 a^2 c x^2}+\frac{b \sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 a^{5/2}}-\frac{\sqrt{c+d x^4}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^4]/(x^7*(a + b*x^4)),x]

[Out]

-Sqrt[c + d*x^4]/(6*a*x^6) + ((3*b*c - a*d)*Sqrt[c + d*x^4])/(6*a^2*c*x^2) + (b*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b
*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(2*a^(5/2))

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 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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^7 \left (a+b x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{c+d x^2}}{x^4 \left (a+b x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{c+d x^4}}{6 a x^6}+\frac{\operatorname{Subst}\left (\int \frac{-3 b c+a d-2 b d x^2}{x^2 \left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^2\right )}{6 a}\\ &=-\frac{\sqrt{c+d x^4}}{6 a x^6}+\frac{(3 b c-a d) \sqrt{c+d x^4}}{6 a^2 c x^2}-\frac{\operatorname{Subst}\left (\int -\frac{3 b c (b c-a d)}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^2\right )}{6 a^2 c}\\ &=-\frac{\sqrt{c+d x^4}}{6 a x^6}+\frac{(3 b c-a d) \sqrt{c+d x^4}}{6 a^2 c x^2}+\frac{(b (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^2}\\ &=-\frac{\sqrt{c+d x^4}}{6 a x^6}+\frac{(3 b c-a d) \sqrt{c+d x^4}}{6 a^2 c x^2}+\frac{(b (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^2}\\ &=-\frac{\sqrt{c+d x^4}}{6 a x^6}+\frac{(3 b c-a d) \sqrt{c+d x^4}}{6 a^2 c x^2}+\frac{b \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^{5/2}}\\ \end{align*}

Mathematica [C]  time = 1.06978, size = 221, normalized size = 2.01 \[ -\frac{\sqrt{c+d x^4} \left (\frac{d x^4}{c}+1\right ) \left (\frac{4 x^4 \left (c+d x^4\right ) (a d-b c) \, _2F_1\left (2,2;\frac{3}{2};\frac{(b c-a d) x^4}{c \left (b x^4+a\right )}\right )}{c^2 \left (a+b x^4\right )}+\frac{\left (c-2 d x^4\right ) \left (\sqrt{\frac{a \left (c+d x^4\right )}{c \left (a+b x^4\right )}}+\sqrt{\frac{x^4 (b c-a d)}{c \left (a+b x^4\right )}} \sin ^{-1}\left (\sqrt{\frac{x^4 (b c-a d)}{c \left (a+b x^4\right )}}\right )\right )}{c \left (\frac{a \left (c+d x^4\right )}{c \left (a+b x^4\right )}\right )^{3/2}}\right )}{6 x^6 \left (a+b x^4\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c + d*x^4]/(x^7*(a + b*x^4)),x]

[Out]

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

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Maple [B]  time = 0.014, size = 1116, normalized size = 10.2 \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^7/(b*x^4+a),x)

[Out]

1/2/a^2*b/c/x^2*(d*x^4+c)^(3/2)-1/2/a^2*b*d/c*x^2*(d*x^4+c)^(1/2)-1/2/a^2*b*d^(1/2)*ln(x^2*d^(1/2)+(d*x^4+c)^(
1/2))+1/4*b^2/a^2/(-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*b/a^2*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*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))*d-1/4*b^2/a^2/(-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^2/a^2/(-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*b/a^2*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*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))*d+1/4*b^2/a^2/(-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-1/6/a*(d*x^4+c)^(3/2)/x^6/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^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^7/(b*x^4+a),x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^4 + c)/((b*x^4 + a)*x^7), x)

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Fricas [A]  time = 2.00939, size = 684, normalized size = 6.22 \begin{align*} \left [\frac{3 \, b c x^{6} \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 \,{\left ({\left (3 \, b c - a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c}}{24 \, a^{2} c x^{6}}, \frac{3 \, b c x^{6} \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 \,{\left ({\left (3 \, b c - a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c}}{12 \, a^{2} c x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^7/(b*x^4+a),x, algorithm="fricas")

[Out]

[1/24*(3*b*c*x^6*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*((3*b*c - a*d)*x^4 - a*c)*sqrt(d*x^4 + c))/(a^2*c*x^6), 1/12*(3*b*c*x^6*sqrt((b*c - a*d)/a)*arct
an(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*((3*b*c - a*d)*x^4 - a*c)*sqrt(d*x^4 + c))/(a^2*c*x^6)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{4}}}{x^{7} \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**7/(b*x**4+a),x)

[Out]

Integral(sqrt(c + d*x**4)/(x**7*(a + b*x**4)), x)

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Giac [A]  time = 1.21403, size = 131, normalized size = 1.19 \begin{align*} -\frac{\frac{3 \,{\left (b^{2} c^{2} - a b c d\right )} \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{4}}}}{\sqrt{a b c - a^{2} d}}\right )}{\sqrt{a b c - a^{2} d} a^{2}} - \frac{3 \, a b c \sqrt{d + \frac{c}{x^{4}}} - a^{2}{\left (d + \frac{c}{x^{4}}\right )}^{\frac{3}{2}}}{a^{3}}}{6 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^4+c)^(1/2)/x^7/(b*x^4+a),x, algorithm="giac")

[Out]

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