3.248 \(\int \frac {A+B x^3}{x^4 (a+b x^3)^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{7/2}}-\frac {5 A b-2 a B}{3 a^3 \sqrt {a+b x^3}}+\frac {2 a B-5 A b}{9 a^2 \left (a+b x^3\right )^{3/2}}-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}} \]

[Out]

1/9*(-5*A*b+2*B*a)/a^2/(b*x^3+a)^(3/2)-1/3*A/a/x^3/(b*x^3+a)^(3/2)+1/3*(5*A*b-2*B*a)*arctanh((b*x^3+a)^(1/2)/a
^(1/2))/a^(7/2)+1/3*(-5*A*b+2*B*a)/a^3/(b*x^3+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 113, 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, 51, 63, 208} \[ -\frac {5 A b-2 a B}{3 a^3 \sqrt {a+b x^3}}-\frac {5 A b-2 a B}{9 a^2 \left (a+b x^3\right )^{3/2}}+\frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{7/2}}-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^4*(a + b*x^3)^(5/2)),x]

[Out]

-(5*A*b - 2*a*B)/(9*a^2*(a + b*x^3)^(3/2)) - A/(3*a*x^3*(a + b*x^3)^(3/2)) - (5*A*b - 2*a*B)/(3*a^3*Sqrt[a + b
*x^3]) + ((5*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(3*a^(7/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {A+B x^3}{x^4 \left (a+b x^3\right )^{5/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx,x,x^3\right )\\ &=-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}}+\frac {\left (-\frac {5 A b}{2}+a B\right ) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,x^3\right )}{3 a}\\ &=-\frac {5 A b-2 a B}{9 a^2 \left (a+b x^3\right )^{3/2}}-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}}-\frac {(5 A b-2 a B) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,x^3\right )}{6 a^2}\\ &=-\frac {5 A b-2 a B}{9 a^2 \left (a+b x^3\right )^{3/2}}-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}}-\frac {5 A b-2 a B}{3 a^3 \sqrt {a+b x^3}}-\frac {(5 A b-2 a B) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{6 a^3}\\ &=-\frac {5 A b-2 a B}{9 a^2 \left (a+b x^3\right )^{3/2}}-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}}-\frac {5 A b-2 a B}{3 a^3 \sqrt {a+b x^3}}-\frac {(5 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 a^3 b}\\ &=-\frac {5 A b-2 a B}{9 a^2 \left (a+b x^3\right )^{3/2}}-\frac {A}{3 a x^3 \left (a+b x^3\right )^{3/2}}-\frac {5 A b-2 a B}{3 a^3 \sqrt {a+b x^3}}+\frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 57, normalized size = 0.50 \[ \frac {x^3 (2 a B-5 A b) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b x^3}{a}+1\right )-3 a A}{9 a^2 x^3 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^4*(a + b*x^3)^(5/2)),x]

[Out]

(-3*a*A + (-5*A*b + 2*a*B)*x^3*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*x^3)/a])/(9*a^2*x^3*(a + b*x^3)^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 351, normalized size = 3.11 \[ \left [-\frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{9} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6} - 3 \, A a^{3} + 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{18 \, {\left (a^{4} b^{2} x^{9} + 2 \, a^{5} b x^{6} + a^{6} x^{3}\right )}}, \frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{9} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6} - 3 \, A a^{3} + 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{9 \, {\left (a^{4} b^{2} x^{9} + 2 \, a^{5} b x^{6} + a^{6} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/18*(3*((2*B*a*b^2 - 5*A*b^3)*x^9 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^6 + (2*B*a^3 - 5*A*a^2*b)*x^3)*sqrt(a)*log(
(b*x^3 + 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) - 2*(3*(2*B*a^2*b - 5*A*a*b^2)*x^6 - 3*A*a^3 + 4*(2*B*a^3 - 5*A
*a^2*b)*x^3)*sqrt(b*x^3 + a))/(a^4*b^2*x^9 + 2*a^5*b*x^6 + a^6*x^3), 1/9*(3*((2*B*a*b^2 - 5*A*b^3)*x^9 + 2*(2*
B*a^2*b - 5*A*a*b^2)*x^6 + (2*B*a^3 - 5*A*a^2*b)*x^3)*sqrt(-a)*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (3*(2*B*a^
2*b - 5*A*a*b^2)*x^6 - 3*A*a^3 + 4*(2*B*a^3 - 5*A*a^2*b)*x^3)*sqrt(b*x^3 + a))/(a^4*b^2*x^9 + 2*a^5*b*x^6 + a^
6*x^3)]

________________________________________________________________________________________

giac [A]  time = 0.19, size = 101, normalized size = 0.89 \[ \frac {{\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a} a^{3}} + \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )} B a + B a^{2} - 6 \, {\left (b x^{3} + a\right )} A b - A a b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x^{3} + a} A}{3 \, a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

1/3*(2*B*a - 5*A*b)*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2/9*(3*(b*x^3 + a)*B*a + B*a^2 - 6*(b*x^
3 + a)*A*b - A*a*b)/((b*x^3 + a)^(3/2)*a^3) - 1/3*sqrt(b*x^3 + a)*A/(a^3*x^3)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 157, normalized size = 1.39 \[ \left (\frac {5 b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {7}{2}}}-\frac {4 b}{3 \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}\, a^{3}}-\frac {2 \sqrt {b \,x^{3}+a}}{9 \left (x^{3}+\frac {a}{b}\right )^{2} a^{2} b}-\frac {\sqrt {b \,x^{3}+a}}{3 a^{3} x^{3}}\right ) A +\left (-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {5}{2}}}+\frac {2}{3 \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}\, a^{2}}+\frac {2 \sqrt {b \,x^{3}+a}}{9 \left (x^{3}+\frac {a}{b}\right )^{2} a \,b^{2}}\right ) B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^4/(b*x^3+a)^(5/2),x)

[Out]

A*(-1/3/a^3*(b*x^3+a)^(1/2)/x^3-2/9/a^2/b*(b*x^3+a)^(1/2)/(x^3+a/b)^2-4/3*b/a^3/((x^3+a/b)*b)^(1/2)+5/3*b/a^(7
/2)*arctanh((b*x^3+a)^(1/2)/a^(1/2)))+B*(2/9*(b*x^3+a)^(1/2)/(x^3+a/b)^2/a/b^2+2/3/((x^3+a/b)*b)^(1/2)/a^2-2/3
/a^(5/2)*arctanh((b*x^3+a)^(1/2)/a^(1/2)))

________________________________________________________________________________________

maxima [A]  time = 1.25, size = 170, normalized size = 1.50 \[ -\frac {1}{18} \, A {\left (\frac {2 \, {\left (15 \, {\left (b x^{3} + a\right )}^{2} b - 10 \, {\left (b x^{3} + a\right )} a b - 2 \, a^{2} b\right )}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3} - {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{4}} + \frac {15 \, b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )} + \frac {1}{9} \, B {\left (\frac {3 \, \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, b x^{3} + 4 \, a\right )}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

-1/18*A*(2*(15*(b*x^3 + a)^2*b - 10*(b*x^3 + a)*a*b - 2*a^2*b)/((b*x^3 + a)^(5/2)*a^3 - (b*x^3 + a)^(3/2)*a^4)
+ 15*b*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(7/2)) + 1/9*B*(3*log((sqrt(b*x^3 + a)
- sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2) + 2*(3*b*x^3 + 4*a)/((b*x^3 + a)^(3/2)*a^2))

________________________________________________________________________________________

mupad [B]  time = 2.97, size = 198, normalized size = 1.75 \[ \frac {\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )\,\left (5\,A\,b-2\,B\,a\right )}{6\,a^{7/2}}-\frac {\frac {2\,B\,a^2-5\,A\,a\,b}{2\,a^4}-\frac {a\,\left (\frac {A\,b^2}{3\,a^4}+\frac {5\,b\,\left (2\,B\,a^2-5\,A\,a\,b\right )}{6\,a^5}\right )}{b}}{\sqrt {b\,x^3+a}}-\frac {\frac {2\,B\,a^3-5\,A\,a^2\,b}{4\,a^4}-\frac {a\,\left (\frac {13\,b\,\left (2\,B\,a^3-5\,A\,a^2\,b\right )}{36\,a^5}+\frac {A\,b^2}{3\,a^3}\right )}{b}}{{\left (b\,x^3+a\right )}^{3/2}}-\frac {A\,\sqrt {b\,x^3+a}}{3\,a^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^4*(a + b*x^3)^(5/2)),x)

[Out]

(log((((a + b*x^3)^(1/2) - a^(1/2))*((a + b*x^3)^(1/2) + a^(1/2))^3)/x^6)*(5*A*b - 2*B*a))/(6*a^(7/2)) - ((2*B
*a^2 - 5*A*a*b)/(2*a^4) - (a*((A*b^2)/(3*a^4) + (5*b*(2*B*a^2 - 5*A*a*b))/(6*a^5)))/b)/(a + b*x^3)^(1/2) - ((2
*B*a^3 - 5*A*a^2*b)/(4*a^4) - (a*((13*b*(2*B*a^3 - 5*A*a^2*b))/(36*a^5) + (A*b^2)/(3*a^3)))/b)/(a + b*x^3)^(3/
2) - (A*(a + b*x^3)^(1/2))/(3*a^3*x^3)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**4/(b*x**3+a)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________