3.2.85 \(\int \frac {(a+b x^3)^{3/2} (A+B x^3)}{x^4} \, dx\)

Optimal. Leaf size=110 \[ \frac {\left (a+b x^3\right )^{3/2} (2 a B+3 A b)}{9 a}+\frac {1}{3} \sqrt {a+b x^3} (2 a B+3 A b)-\frac {1}{3} \sqrt {a} (2 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 78, 50, 63, 208} \begin {gather*} \frac {\left (a+b x^3\right )^{3/2} (2 a B+3 A b)}{9 a}+\frac {1}{3} \sqrt {a+b x^3} (2 a B+3 A b)-\frac {1}{3} \sqrt {a} (2 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^(3/2)*(A + B*x^3))/x^4,x]

[Out]

((3*A*b + 2*a*B)*Sqrt[a + b*x^3])/3 + ((3*A*b + 2*a*B)*(a + b*x^3)^(3/2))/(9*a) - (A*(a + b*x^3)^(5/2))/(3*a*x
^3) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/3

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 78

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

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 {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^4} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx,x,x^3\right )\\ &=-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac {\left (\frac {3 A b}{2}+a B\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^3\right )}{3 a}\\ &=\frac {(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac {1}{6} (3 A b+2 a B) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^3\right )\\ &=\frac {1}{3} (3 A b+2 a B) \sqrt {a+b x^3}+\frac {(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac {1}{6} (a (3 A b+2 a B)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=\frac {1}{3} (3 A b+2 a B) \sqrt {a+b x^3}+\frac {(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac {(a (3 A b+2 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{3 b}\\ &=\frac {1}{3} (3 A b+2 a B) \sqrt {a+b x^3}+\frac {(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac {A \left (a+b x^3\right )^{5/2}}{3 a x^3}-\frac {1}{3} \sqrt {a} (3 A b+2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 80, normalized size = 0.73 \begin {gather*} \frac {1}{9} \left (\frac {\sqrt {a+b x^3} \left (-3 a A+8 a B x^3+6 A b x^3+2 b B x^6\right )}{x^3}-3 \sqrt {a} (2 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^(3/2)*(A + B*x^3))/x^4,x]

[Out]

((Sqrt[a + b*x^3]*(-3*a*A + 6*A*b*x^3 + 8*a*B*x^3 + 2*b*B*x^6))/x^3 - 3*Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a
+ b*x^3]/Sqrt[a]])/9

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IntegrateAlgebraic [A]  time = 0.11, size = 85, normalized size = 0.77 \begin {gather*} \frac {1}{3} \left (-2 a^{3/2} B-3 \sqrt {a} A b\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )+\frac {\sqrt {a+b x^3} \left (-3 a A+8 a B x^3+6 A b x^3+2 b B x^6\right )}{9 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^3)^(3/2)*(A + B*x^3))/x^4,x]

[Out]

(Sqrt[a + b*x^3]*(-3*a*A + 6*A*b*x^3 + 8*a*B*x^3 + 2*b*B*x^6))/(9*x^3) + ((-3*Sqrt[a]*A*b - 2*a^(3/2)*B)*ArcTa
nh[Sqrt[a + b*x^3]/Sqrt[a]])/3

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fricas [A]  time = 0.57, size = 169, normalized size = 1.54 \begin {gather*} \left [\frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (2 \, B b x^{6} + 2 \, {\left (4 \, B a + 3 \, A b\right )} x^{3} - 3 \, A a\right )} \sqrt {b x^{3} + a}}{18 \, x^{3}}, \frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (2 \, B b x^{6} + 2 \, {\left (4 \, B a + 3 \, A b\right )} x^{3} - 3 \, A a\right )} \sqrt {b x^{3} + a}}{9 \, x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x, algorithm="fricas")

[Out]

[1/18*(3*(2*B*a + 3*A*b)*sqrt(a)*x^3*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(2*B*b*x^6 + 2*(4*
B*a + 3*A*b)*x^3 - 3*A*a)*sqrt(b*x^3 + a))/x^3, 1/9*(3*(2*B*a + 3*A*b)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a)*sqr
t(-a)/a) + (2*B*b*x^6 + 2*(4*B*a + 3*A*b)*x^3 - 3*A*a)*sqrt(b*x^3 + a))/x^3]

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giac [A]  time = 0.20, size = 103, normalized size = 0.94 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B b + 6 \, \sqrt {b x^{3} + a} B a b + 6 \, \sqrt {b x^{3} + a} A b^{2} + \frac {3 \, {\left (2 \, B a^{2} b + 3 \, A a b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, \sqrt {b x^{3} + a} A a b}{x^{3}}}{9 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x, algorithm="giac")

[Out]

1/9*(2*(b*x^3 + a)^(3/2)*B*b + 6*sqrt(b*x^3 + a)*B*a*b + 6*sqrt(b*x^3 + a)*A*b^2 + 3*(2*B*a^2*b + 3*A*a*b^2)*a
rctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) - 3*sqrt(b*x^3 + a)*A*a*b/x^3)/b

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maple [A]  time = 0.05, size = 101, normalized size = 0.92 \begin {gather*} \left (-\sqrt {a}\, b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b \,x^{3}+a}\, b}{3}-\frac {\sqrt {b \,x^{3}+a}\, a}{3 x^{3}}\right ) A +\left (\frac {2 \sqrt {b \,x^{3}+a}\, b \,x^{3}}{9}-\frac {2 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3}+\frac {8 \sqrt {b \,x^{3}+a}\, a}{9}\right ) B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x)

[Out]

A*(-1/3*(b*x^3+a)^(1/2)*a/x^3+2/3*(b*x^3+a)^(1/2)*b-b*arctanh((b*x^3+a)^(1/2)/a^(1/2))*a^(1/2))+B*(2/9*(b*x^3+
a)^(1/2)*b*x^3+8/9*(b*x^3+a)^(1/2)*a-2/3*a^(3/2)*arctanh((b*x^3+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 1.44, size = 134, normalized size = 1.22 \begin {gather*} \frac {1}{6} \, {\left (3 \, \sqrt {a} b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + 4 \, \sqrt {b x^{3} + a} b - \frac {2 \, \sqrt {b x^{3} + a} a}{x^{3}}\right )} A + \frac {1}{9} \, {\left (3 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + 2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} + 6 \, \sqrt {b x^{3} + a} a\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x, algorithm="maxima")

[Out]

1/6*(3*sqrt(a)*b*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a))) + 4*sqrt(b*x^3 + a)*b - 2*sqrt(b
*x^3 + a)*a/x^3)*A + 1/9*(3*a^(3/2)*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a))) + 2*(b*x^3 +
a)^(3/2) + 6*sqrt(b*x^3 + a)*a)*B

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mupad [B]  time = 3.38, size = 111, normalized size = 1.01 \begin {gather*} \frac {\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {\frac {a}{4}}}{3}+\frac {\left (2\,A\,b^2+\frac {8\,B\,a\,b}{3}\right )\,\sqrt {b\,x^3+a}}{3\,b}-\frac {A\,a\,\sqrt {b\,x^3+a}}{3\,x^3}+\frac {2\,B\,b\,x^3\,\sqrt {b\,x^3+a}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^(3/2))/x^4,x)

[Out]

(log((((a + b*x^3)^(1/2) - a^(1/2))^3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6)*(3*A*b + 2*B*a)*(a/4)^(1/2))/3 + ((2
*A*b^2 + (8*B*a*b)/3)*(a + b*x^3)^(1/2))/(3*b) - (A*a*(a + b*x^3)^(1/2))/(3*x^3) + (2*B*b*x^3*(a + b*x^3)^(1/2
))/9

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sympy [A]  time = 58.40, size = 223, normalized size = 2.03 \begin {gather*} - A \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 x^{\frac {3}{2}}} + \frac {2 A a \sqrt {b}}{3 x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {2 A b^{\frac {3}{2}} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {2 B a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3} + \frac {2 B a^{2}}{3 \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {2 B a \sqrt {b} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x^{3}} + 1}} + B b \left (\begin {cases} \frac {\sqrt {a} x^{3}}{3} & \text {for}\: b = 0 \\\frac {2 \left (a + b x^{3}\right )^{\frac {3}{2}}}{9 b} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(3/2)*(B*x**3+A)/x**4,x)

[Out]

-A*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x**(3/2))) - A*a*sqrt(b)*sqrt(a/(b*x**3) + 1)/(3*x**(3/2)) + 2*A*a*sqrt(b)
/(3*x**(3/2)*sqrt(a/(b*x**3) + 1)) + 2*A*b**(3/2)*x**(3/2)/(3*sqrt(a/(b*x**3) + 1)) - 2*B*a**(3/2)*asinh(sqrt(
a)/(sqrt(b)*x**(3/2)))/3 + 2*B*a**2/(3*sqrt(b)*x**(3/2)*sqrt(a/(b*x**3) + 1)) + 2*B*a*sqrt(b)*x**(3/2)/(3*sqrt
(a/(b*x**3) + 1)) + B*b*Piecewise((sqrt(a)*x**3/3, Eq(b, 0)), (2*(a + b*x**3)**(3/2)/(9*b), True))

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