3.5.72 \(\int \frac {x^2 (c+d x^3)^{3/2}}{(a+b x^3)^2} \, dx\) [472]

Optimal. Leaf size=94 \[ \frac {d \sqrt {c+d x^3}}{b^2}-\frac {\left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right )}-\frac {d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {455, 43, 52, 65, 214} \begin {gather*} -\frac {d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-\frac {\left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right )}+\frac {d \sqrt {c+d x^3}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 65

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 214

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.20, size = 94, normalized size = 1.00 \begin {gather*} \frac {\sqrt {c+d x^3} \left (-b c+3 a d+2 b d x^3\right )}{3 b^2 \left (a+b x^3\right )}-\frac {d \sqrt {-b c+a d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.38, size = 466, normalized size = 4.96

method	result	size
default	\(\frac {\left (a d -b c \right ) \sqrt {d \,x^{3}+c}}{3 b^{2} \left (b \,x^{3}+a \right )}+\frac {2 d \sqrt {d \,x^{3}+c}}{3 b^{2}}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\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 {b \left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d \right )}{2 d \left (a d -b c \right )}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{2 d \,b^{2}}\)	\(466\)
elliptic	\(\frac {\left (a d -b c \right ) \sqrt {d \,x^{3}+c}}{3 b^{2} \left (b \,x^{3}+a \right )}+\frac {2 d \sqrt {d \,x^{3}+c}}{3 b^{2}}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\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 {b \left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d \right )}{2 d \left (a d -b c \right )}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{2 d \,b^{2}}\)	\(466\)
risch	\(\text {Expression too large to display}\)	\(936\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Fricas [A]
time = 1.59, size = 234, normalized size = 2.49 \begin {gather*} \left [\frac {3 \, {\left (b d x^{3} + 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 ) + 2 \, {\left (2 \, b d x^{3} - b c + 3 \, a d\right )} \sqrt {d x^{3} + c}}{6 \, {\left (b^{3} x^{3} + a b^{2}\right )}}, -\frac {3 \, {\left (b d x^{3} + 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 ) - {\left (2 \, b d x^{3} - b c + 3 \, a d\right )} \sqrt {d x^{3} + c}}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (a + b x^{3}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.21, size = 122, normalized size = 1.30 \begin {gather*} \frac {2 \, \sqrt {d x^{3} + c} d}{3 \, b^{2}} + \frac {{\left (b c d - a 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} b^{2}} - \frac {\sqrt {d x^{3} + c} b c d - \sqrt {d x^{3} + c} a d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 7.35, size = 170, normalized size = 1.81 \begin {gather*} \frac {2\,d\,\sqrt {d\,x^3+c}}{3\,b^2}-\frac {\left (\frac {2\,b\,c^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {a\,\left (\frac {2\,a\,d^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {4\,b\,c\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )}{b}\right )\,\sqrt {d\,x^3+c}}{b\,x^3+a}+\frac {d\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,1{}\mathrm {i}}{2\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d*(c + d*x^3)^(1/2))/(3*b^2) - (((2*b*c^2)/(3*(2*b^2*c - 2*a*b*d)) + (a*((2*a*d^2)/(3*(2*b^2*c - 2*a*b*d))
- (4*b*c*d)/(3*(2*b^2*c - 2*a*b*d))))/b)*(c + d*x^3)^(1/2))/(a + b*x^3) + (d*log((a*d - 2*b*c + b^(1/2)*(c + d
*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i - b*d*x^3)/(a + b*x^3))*(a*d - b*c)^(1/2)*1i)/(2*b^(5/2))

________________________________________________________________________________________