3.382 \(\int \frac{1}{x^4 (a+b x^3) \sqrt{c+d x^3}} \, dx\)

Optimal. Leaf size=117 \[ -\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{3/2}}-\frac{\sqrt{c+d x^3}}{3 a c x^3} \]

[Out]

-Sqrt[c + d*x^3]/(3*a*c*x^3) + ((2*b*c + a*d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(3/2)) - (2*b^(3/2)*A
rcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*a^2*Sqrt[b*c - a*d])

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Rubi [A]  time = 0.118177, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 103, 156, 63, 208} \[ -\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{3/2}}-\frac{\sqrt{c+d x^3}}{3 a c x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^3)*Sqrt[c + d*x^3]),x]

[Out]

-Sqrt[c + d*x^3]/(3*a*c*x^3) + ((2*b*c + a*d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(3/2)) - (2*b^(3/2)*A
rcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*a^2*Sqrt[b*c - a*d])

Rule 446

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.112794, size = 151, normalized size = 1.29 \[ \frac{2 b^{3/2} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (a d-b c)}+\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 \sqrt{c}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a c^{3/2}}-\frac{\sqrt{c+d x^3}}{3 a c x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^3)*Sqrt[c + d*x^3]),x]

[Out]

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

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Maple [C]  time = 0.01, size = 498, normalized size = 4.3 \begin{align*}{\frac{-{\frac{i}{3}}{b}^{2}\sqrt{2}}{{a}^{2}{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\frac{1}{ad-bc}\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}d \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},{\frac{b}{2\,d \left ( ad-bc \right ) } \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}}+{\frac{1}{a} \left ( -{\frac{1}{3\,c{x}^{3}}\sqrt{d{x}^{3}+c}}+{\frac{d}{3}{\it Artanh} \left ({\sqrt{d{x}^{3}+c}{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \right ) }+{\frac{2\,b}{3\,{a}^{2}}{\it Artanh} \left ({\sqrt{d{x}^{3}+c}{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*b^2/a^2/d^2*2^(1/2)*sum(1/(a*d-b*c)*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c
)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*
(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c
)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*Ellip
ticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1
/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_
alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/
2)),_alpha=RootOf(_Z^3*b+a))+1/a*(-1/3*(d*x^3+c)^(1/2)/c/x^3+1/3*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))+2
/3/a^2*b*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )} \sqrt{d x^{3} + c} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.80769, size = 1261, normalized size = 10.78 \begin{align*} \left [\frac{2 \, b c^{2} x^{3} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a d}}}{b x^{3} + a}\right ) +{\left (2 \, b c + a d\right )} \sqrt{c} x^{3} \log \left (\frac{d x^{3} + 2 \, \sqrt{d x^{3} + c} \sqrt{c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt{d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}, -\frac{4 \, b c^{2} x^{3} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{3} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a d}}}{b d x^{3} + b c}\right ) -{\left (2 \, b c + a d\right )} \sqrt{c} x^{3} \log \left (\frac{d x^{3} + 2 \, \sqrt{d x^{3} + c} \sqrt{c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt{d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}, \frac{b c^{2} x^{3} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a d}}}{b x^{3} + a}\right ) -{\left (2 \, b c + a d\right )} \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) - \sqrt{d x^{3} + c} a c}{3 \, a^{2} c^{2} x^{3}}, -\frac{2 \, b c^{2} x^{3} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{3} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a d}}}{b d x^{3} + b c}\right ) +{\left (2 \, b c + a d\right )} \sqrt{-c} x^{3} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) + \sqrt{d x^{3} + c} a c}{3 \, a^{2} c^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (a + b x^{3}\right ) \sqrt{c + d x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**4*(a + b*x**3)*sqrt(c + d*x**3)), x)

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Giac [A]  time = 1.11741, size = 159, normalized size = 1.36 \begin{align*} \frac{1}{3} \, d^{2}{\left (\frac{2 \, b^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{2} d^{2}} - \frac{{\left (2 \, b c + a d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c d^{2}} - \frac{\sqrt{d x^{3} + c}}{a c d^{2} x^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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