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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0869774, antiderivative size = 113, normalized size of antiderivative = 1., 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 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 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 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 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{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 = 0.0203189, size = 57, normalized size = 0.5 \[ \frac{x^3 (2 a B-5 A b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b x^3}{a}+1\right )-3 a A}{9 a^2 x^3 \left (a+b x^3\right )^{3/2}} \]

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))

________________________________________________________________________________________

Maple [A]  time = 0.028, size = 157, normalized size = 1.4 \begin{align*} A \left ( -{\frac{2}{9\,{a}^{2}b}\sqrt{b{x}^{3}+a} \left ({x}^{3}+{\frac{a}{b}} \right ) ^{-2}}-{\frac{4\,b}{3\,{a}^{3}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{1}{3\,{a}^{3}{x}^{3}}\sqrt{b{x}^{3}+a}}+{\frac{5\,b}{3}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{7}{2}}}} \right ) +B \left ({\frac{2}{9\,a{b}^{2}}\sqrt{b{x}^{3}+a} \left ({x}^{3}+{\frac{a}{b}} \right ) ^{-2}}+{\frac{2}{3\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{2}{3}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}} \right ) \end{align*}

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.77859, size = 757, normalized size = 6.7 \begin{align*} \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 ] \end{align*}

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)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.17357, size = 136, normalized size = 1.2 \begin{align*} \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}} \end{align*}

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)