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

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

[Out]

-2/9*a*(d*x^3+c)^(3/2)/b^2+2/15*(d*x^3+c)^(5/2)/b/d+2/3*a*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a
*d+b*c)^(1/2))/b^(7/2)-2/3*a*(-a*d+b*c)*(d*x^3+c)^(1/2)/b^3

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, 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 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 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 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]

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]]

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 = 0.12, 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))

________________________________________________________________________________________

fricas [A]  time = 1.76, size = 297, normalized size = 2.48 \[ \left [-\frac {15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{6} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{45 \, b^{3} d}, \frac {2 \, {\left (15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (3 \, b^{2} d^{2} x^{6} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, b^{3} d}\right ] \]

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)]

________________________________________________________________________________________

giac [A]  time = 0.19, size = 151, normalized size = 1.26 \[ -\frac {2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{4} d^{4} - 5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a b^{3} d^{5} - 15 \, \sqrt {d x^{3} + c} a b^{3} c d^{5} + 15 \, \sqrt {d x^{3} + c} a^{2} b^{2} d^{6}\right )}}{45 \, b^{5} d^{5}} \]

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)

________________________________________________________________________________________

maple [C]  time = 0.32, size = 531, normalized size = 4.42 \[ -\frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, d \,x^{3}}{9 b}+\frac {2 \left (-\frac {2 c d}{3 b}-\frac {\left (a d -2 b c \right ) d}{b^{2}}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \left (-a^{2} d^{2}+2 a b c d -b^{2} c^{2}\right ) \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {\left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right )\right ) b}{2 \left (a d -b c \right ) d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 b^{2} d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right ) a}{b}+\frac {2 \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 b d} \]

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*x^3+c)^(1/2)/b*d*x^3+2/3*(-2/3/b*c*d-(a*d-2*b*c)/b^2*d)*(d*x^3+c)^(1/2)/d
+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)*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d
^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^
2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)
^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*
d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)
^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha
-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2
)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a)))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

________________________________________________________________________________________

mupad [B]  time = 6.13, size = 215, normalized size = 1.79 \[ \frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,c^2}{b}+\frac {2\,a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}+\frac {2\,c\,\left (\frac {2\,a\,d^2}{b^2}-\frac {12\,c\,d}{5\,b}\right )}{3\,d}\right )}{3\,d}+\frac {2\,d\,x^6\,\sqrt {d\,x^3+c}}{15\,b}-\frac {x^3\,\sqrt {d\,x^3+c}\,\left (\frac {2\,a\,d^2}{b^2}-\frac {12\,c\,d}{5\,b}\right )}{9\,d}+\frac {a\,\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d+\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*x^3)^(1/2)*((2*c^2)/b + (2*a*((a*d^2)/b^2 - (2*c*d)/b))/b + (2*c*((2*a*d^2)/b^2 - (12*c*d)/(5*b)))/(3*
d)))/(3*d) + (2*d*x^6*(c + d*x^3)^(1/2))/(15*b) - (x^3*(c + d*x^3)^(1/2)*((2*a*d^2)/b^2 - (12*c*d)/(5*b)))/(9*
d) + (a*log((a^2*d^2 + 2*b^2*c^2 + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(3/2)*2i - a*b*d^2*x^3 + b^2*c*d*x^3
- 3*a*b*c*d)/(a + b*x^3))*(a*d - b*c)^(3/2)*1i)/(3*b^(7/2))

________________________________________________________________________________________

sympy [A]  time = 70.60, size = 116, normalized size = 0.97 \[ - \frac {2 a \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b^{2}} - \frac {2 a \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{4} \sqrt {\frac {a d - b c}{b}}} + \frac {2 \left (c + d x^{3}\right )^{\frac {5}{2}}}{15 b d} + \frac {\sqrt {c + d x^{3}} \left (2 a^{2} d - 2 a b c\right )}{3 b^{3}} \]

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)

________________________________________________________________________________________