3.233 \(\int \frac{A+B x^3}{x^7 (a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=118 \[ \frac{b (5 A b-4 a B)}{4 a^3 \sqrt{a+b x^3}}+\frac{5 A b-4 a B}{12 a^2 x^3 \sqrt{a+b x^3}}-\frac{b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{A}{6 a x^6 \sqrt{a+b x^3}} \]

[Out]

(b*(5*A*b - 4*a*B))/(4*a^3*Sqrt[a + b*x^3]) - A/(6*a*x^6*Sqrt[a + b*x^3]) + (5*A*b - 4*a*B)/(12*a^2*x^3*Sqrt[a
+ b*x^3]) - (b*(5*A*b - 4*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(7/2))

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Rubi [A]  time = 0.0899382, antiderivative size = 120, normalized size of antiderivative = 1.02, 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} \[ \frac{\sqrt{a+b x^3} (5 A b-4 a B)}{4 a^3 x^3}-\frac{5 A b-4 a B}{6 a^2 x^3 \sqrt{a+b x^3}}-\frac{b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{A}{6 a x^6 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^7*(a + b*x^3)^(3/2)),x]

[Out]

-A/(6*a*x^6*Sqrt[a + b*x^3]) - (5*A*b - 4*a*B)/(6*a^2*x^3*Sqrt[a + b*x^3]) + ((5*A*b - 4*a*B)*Sqrt[a + b*x^3])
/(4*a^3*x^3) - (b*(5*A*b - 4*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*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^7 \left (a+b x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 (a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^3}}+\frac{\left (-\frac{5 A b}{2}+2 a B\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,x^3\right )}{6 a}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^3}}-\frac{5 A b-4 a B}{6 a^2 x^3 \sqrt{a+b x^3}}-\frac{(5 A b-4 a B) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^3\right )}{4 a^2}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^3}}-\frac{5 A b-4 a B}{6 a^2 x^3 \sqrt{a+b x^3}}+\frac{(5 A b-4 a B) \sqrt{a+b x^3}}{4 a^3 x^3}+\frac{(b (5 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )}{8 a^3}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^3}}-\frac{5 A b-4 a B}{6 a^2 x^3 \sqrt{a+b x^3}}+\frac{(5 A b-4 a B) \sqrt{a+b x^3}}{4 a^3 x^3}+\frac{(5 A b-4 a B) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^3}\right )}{4 a^3}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^3}}-\frac{5 A b-4 a B}{6 a^2 x^3 \sqrt{a+b x^3}}+\frac{(5 A b-4 a B) \sqrt{a+b x^3}}{4 a^3 x^3}-\frac{b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0199701, size = 60, normalized size = 0.51 \[ \frac{b x^6 (5 A b-4 a B) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x^3}{a}+1\right )-a^2 A}{6 a^3 x^6 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^7*(a + b*x^3)^(3/2)),x]

[Out]

(-(a^2*A) + b*(5*A*b - 4*a*B)*x^6*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (b*x^3)/a])/(6*a^3*x^6*Sqrt[a + b*x^3])

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Maple [A]  time = 0.026, size = 141, normalized size = 1.2 \begin{align*} A \left ({\frac{2\,{b}^{2}}{3\,{a}^{3}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{1}{6\,{a}^{2}{x}^{6}}\sqrt{b{x}^{3}+a}}+{\frac{7\,b}{12\,{a}^{3}{x}^{3}}\sqrt{b{x}^{3}+a}}-{\frac{5\,{b}^{2}}{4}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{7}{2}}}} \right ) +B \left ( -{\frac{2\,b}{3\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{1}{3\,{a}^{2}{x}^{3}}\sqrt{b{x}^{3}+a}}+{b{\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^7/(b*x^3+a)^(3/2),x)

[Out]

A*(2/3*b^2/a^3/((x^3+a/b)*b)^(1/2)-1/6/a^2*(b*x^3+a)^(1/2)/x^6+7/12*b/a^3*(b*x^3+a)^(1/2)/x^3-5/4/a^(7/2)*b^2*
arctanh((b*x^3+a)^(1/2)/a^(1/2)))+B*(-2/3*b/a^2/((x^3+a/b)*b)^(1/2)-1/3/a^2*(b*x^3+a)^(1/2)/x^3+1/a^(5/2)*b*ar
ctanh((b*x^3+a)^(1/2)/a^(1/2)))

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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^7/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79482, size = 630, normalized size = 5.34 \begin{align*} \left [-\frac{3 \,{\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{9} +{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6}\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 (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6} + 2 \, A a^{3} +{\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{3}\right )} \sqrt{b x^{3} + a}}{24 \,{\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}, -\frac{3 \,{\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{9} +{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (3 \,{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{6} + 2 \, A a^{3} +{\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{3}\right )} \sqrt{b x^{3} + a}}{12 \,{\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^7/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/24*(3*((4*B*a*b^2 - 5*A*b^3)*x^9 + (4*B*a^2*b - 5*A*a*b^2)*x^6)*sqrt(a)*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqr
t(a) + 2*a)/x^3) + 2*(3*(4*B*a^2*b - 5*A*a*b^2)*x^6 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^3)*sqrt(b*x^3 + a))/(a
^4*b*x^9 + a^5*x^6), -1/12*(3*((4*B*a*b^2 - 5*A*b^3)*x^9 + (4*B*a^2*b - 5*A*a*b^2)*x^6)*sqrt(-a)*arctan(sqrt(b
*x^3 + a)*sqrt(-a)/a) + (3*(4*B*a^2*b - 5*A*a*b^2)*x^6 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^3)*sqrt(b*x^3 + a))
/(a^4*b*x^9 + a^5*x^6)]

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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**7/(b*x**3+a)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1264, size = 185, normalized size = 1.57 \begin{align*} -\frac{{\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a} a^{3}} - \frac{2 \,{\left (B a b - A b^{2}\right )}}{3 \, \sqrt{b x^{3} + a} a^{3}} - \frac{4 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} B a b - 4 \, \sqrt{b x^{3} + a} B a^{2} b - 7 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} A b^{2} + 9 \, \sqrt{b x^{3} + a} A a b^{2}}{12 \, a^{3} b^{2} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^7/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

-1/4*(4*B*a*b - 5*A*b^2)*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3*(B*a*b - A*b^2)/(sqrt(b*x^3 + a
)*a^3) - 1/12*(4*(b*x^3 + a)^(3/2)*B*a*b - 4*sqrt(b*x^3 + a)*B*a^2*b - 7*(b*x^3 + a)^(3/2)*A*b^2 + 9*sqrt(b*x^
3 + a)*A*a*b^2)/(a^3*b^2*x^6)