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

Optimal. Leaf size=120 \[ -\frac{2 a \left (c+d x^3\right )^{3/2}}{9 b^2}-\frac{2 a \sqrt{c+d x^3} (b c-a d)}{3 b^3}+\frac{2 a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d} \]

[Out]

(-2*a*(b*c - a*d)*Sqrt[c + d*x^3])/(3*b^3) - (2*a*(c + d*x^3)^(3/2))/(9*b^2) + (2*(c + d*x^3)^(5/2))/(15*b*d)
+ (2*a*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(7/2))

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Rubi [A]  time = 0.103848, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 80, 50, 63, 208} \[ -\frac{2 a \left (c+d x^3\right )^{3/2}}{9 b^2}-\frac{2 a \sqrt{c+d x^3} (b c-a d)}{3 b^3}+\frac{2 a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(b*c - a*d)*Sqrt[c + d*x^3])/(3*b^3) - (2*a*(c + d*x^3)^(3/2))/(9*b^2) + (2*(c + d*x^3)^(5/2))/(15*b*d)
+ (2*a*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(7/2))

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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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{x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x (c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )\\ &=\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d}-\frac{a \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )}{3 b}\\ &=-\frac{2 a \left (c+d x^3\right )^{3/2}}{9 b^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d}-\frac{(a (b c-a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )}{3 b^2}\\ &=-\frac{2 a (b c-a d) \sqrt{c+d x^3}}{3 b^3}-\frac{2 a \left (c+d x^3\right )^{3/2}}{9 b^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d}-\frac{\left (a (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 b^3}\\ &=-\frac{2 a (b c-a d) \sqrt{c+d x^3}}{3 b^3}-\frac{2 a \left (c+d x^3\right )^{3/2}}{9 b^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d}-\frac{\left (2 a (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 b^3 d}\\ &=-\frac{2 a (b c-a d) \sqrt{c+d x^3}}{3 b^3}-\frac{2 a \left (c+d x^3\right )^{3/2}}{9 b^2}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 b d}+\frac{2 a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.110194, size = 111, normalized size = 0.92 \[ \frac{2 \sqrt{c+d x^3} \left (15 a^2 d^2-5 a b d \left (4 c+d x^3\right )+3 b^2 \left (c+d x^3\right )^2\right )}{45 b^3 d}+\frac{2 a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x^3]*(15*a^2*d^2 + 3*b^2*(c + d*x^3)^2 - 5*a*b*d*(4*c + d*x^3)))/(45*b^3*d) + (2*a*(b*c - a*d)^(
3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(7/2))

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Maple [C]  time = 0.009, size = 531, normalized size = 4.4 \begin{align*}{\frac{2}{15\,bd} \left ( d{x}^{3}+c \right ) ^{{\frac{5}{2}}}}-{\frac{a}{b} \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(d*x^3+c)^(5/2)/b/d-a/b*(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)*sum((-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))*EllipticPi(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)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97544, size = 639, normalized size = 5.32 \begin{align*} \left [-\frac{15 \,{\left (a b c d - a^{2} d^{2}\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 ) - 2 \,{\left (3 \, b^{2} d^{2} x^{6} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} +{\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{45 \, b^{3} d}, \frac{2 \,{\left (15 \,{\left (a b c d - a^{2} d^{2}\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 ) +{\left (3 \, b^{2} d^{2} x^{6} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} +{\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, b^{3} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/45*(15*(a*b*c*d - a^2*d^2)*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)) - 2*(3*b^2*d^2*x^6 + 3*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^3
)*sqrt(d*x^3 + c))/(b^3*d), 2/45*(15*(a*b*c*d - a^2*d^2)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-
(b*c - a*d)/b)/(b*c - a*d)) + (3*b^2*d^2*x^6 + 3*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x
^3)*sqrt(d*x^3 + c))/(b^3*d)]

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Sympy [A]  time = 94.5664, size = 116, normalized size = 0.97 \begin{align*} - \frac{2 a \left (c + d x^{3}\right )^{\frac{3}{2}}}{9 b^{2}} - \frac{2 a \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 b^{4} \sqrt{\frac{a d - b c}{b}}} + \frac{2 \left (c + d x^{3}\right )^{\frac{5}{2}}}{15 b d} + \frac{\sqrt{c + d x^{3}} \left (2 a^{2} d - 2 a b c\right )}{3 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*(c + d*x**3)**(3/2)/(9*b**2) - 2*a*(a*d - b*c)**2*atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b**4*sqrt
((a*d - b*c)/b)) + 2*(c + d*x**3)**(5/2)/(15*b*d) + sqrt(c + d*x**3)*(2*a**2*d - 2*a*b*c)/(3*b**3)

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Giac [A]  time = 1.11057, size = 204, normalized size = 1.7 \begin{align*} -\frac{2 \,{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \, \sqrt{-b^{2} c + a b d} b^{3}} + \frac{2 \,{\left (3 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} b^{4} d^{4} - 5 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} a b^{3} d^{5} - 15 \, \sqrt{d x^{3} + c} a b^{3} c d^{5} + 15 \, \sqrt{d x^{3} + c} a^{2} b^{2} d^{6}\right )}}{45 \, b^{5} d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*
b^3) + 2/45*(3*(d*x^3 + c)^(5/2)*b^4*d^4 - 5*(d*x^3 + c)^(3/2)*a*b^3*d^5 - 15*sqrt(d*x^3 + c)*a*b^3*c*d^5 + 15
*sqrt(d*x^3 + c)*a^2*b^2*d^6)/(b^5*d^5)