3.371 \(\int \frac{(c+d x^3)^{3/2}}{x (a+b x^3)} \, dx\)

Optimal. Leaf size=104 \[ \frac{2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a b^{3/2}}-\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a}+\frac{2 d \sqrt{c+d x^3}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.112071, antiderivative size = 104, 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, 84, 156, 63, 208} \[ \frac{2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a b^{3/2}}-\frac{2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a}+\frac{2 d \sqrt{c+d x^3}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 84

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

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

Mathematica [A]  time = 0.0732751, size = 105, normalized size = 1.01 \[ \frac{2 \left (a \sqrt{b} d \sqrt{c+d x^3}+(b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )-b^{3/2} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )\right )}{3 a b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.009, size = 565, normalized size = 5.4 \begin{align*} -{\frac{b}{a} \left ({\frac{2\,d{x}^{3}}{9\,b}\sqrt{d{x}^{3}+c}}+{\frac{2}{3\,d} \left ( -{\frac{d \left ( ad-2\,bc \right ) }{{b}^{2}}}-{\frac{2\,cd}{3\,b}} \right ) \sqrt{d{x}^{3}+c}}+{\frac{{\frac{i}{3}}\sqrt{2}}{{b}^{2}{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\frac{-{a}^{2}{d}^{2}+2\,abcd-{b}^{2}{c}^{2}}{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}}}}} \right ) }+{\frac{1}{a} \left ({\frac{2\,d{x}^{3}}{9}\sqrt{d{x}^{3}+c}}+{\frac{8\,c}{9}\sqrt{d{x}^{3}+c}}-{\frac{2}{3}{c}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{d{x}^{3}+c}{\frac{1}{\sqrt{c}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/a*(2/9*d/b*x^3*(d*x^3+c)^(1/2)+2/3*(-d*(a*d-2*b*c)/b^2-2/3*d/b*c)/d*(d*x^3+c)^(1/2)+1/3*I/b^2/d^2*2^(1/2)*s
um((-a^2*d^2+2*a*b*c*d-b^2*c^2)/(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))*Ellipt
icPi(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)*_a
lpha-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*(2/9*d*x^3*(d*x^3+c)^(1/2)+8/9*c*(d*x^3+c)^(1/2)-2/3*c^(3/2)*arctanh((d*x^3+c
)^(1/2)/c^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}}}{{\left (b x^{3} + a\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.10785, size = 1080, normalized size = 10.38 \begin{align*} \left [\frac{b c^{\frac{3}{2}} \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 d -{\left (b c - a d\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right )}{3 \, a b}, \frac{b c^{\frac{3}{2}} \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 d + 2 \,{\left (b c - a d\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right )}{3 \, a b}, \frac{2 \, b \sqrt{-c} c \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) + 2 \, \sqrt{d x^{3} + c} a d -{\left (b c - a d\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right )}{3 \, a b}, \frac{2 \,{\left (b \sqrt{-c} c \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) + \sqrt{d x^{3} + c} a d +{\left (b c - a d\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right )\right )}}{3 \, a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 30.248, size = 102, normalized size = 0.98 \begin{align*} \frac{2 d \sqrt{c + d x^{3}}}{3 b} + \frac{2 c^{2} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{- c}} \right )}}{3 a \sqrt{- c}} - \frac{2 \left (a d - b c\right )^{2} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{3 a b^{2} \sqrt{\frac{a d - b c}{b}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*sqrt(c + d*x**3)/(3*b) + 2*c**2*atan(sqrt(c + d*x**3)/sqrt(-c))/(3*a*sqrt(-c)) - 2*(a*d - b*c)**2*atan(sqr
t(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*a*b**2*sqrt((a*d - b*c)/b))

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Giac [A]  time = 1.13406, size = 159, normalized size = 1.53 \begin{align*} \frac{2}{3} \, d{\left (\frac{c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{a \sqrt{-c} d} + \frac{\sqrt{d x^{3} + c}}{b} - \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \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 b d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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