3.2.80 \(\int \frac {\sqrt {a+b x^3} (A+B x^3)}{x^7} \, dx\)

Optimal. Leaf size=88 \[ \frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}}+\frac {\sqrt {a+b x^3} (A b-4 a B)}{12 a x^3}-\frac {A \left (a+b x^3\right )^{3/2}}{6 a x^6} \]

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 78, 47, 63, 208} \begin {gather*} \frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}}+\frac {\sqrt {a+b x^3} (A b-4 a B)}{12 a x^3}-\frac {A \left (a+b x^3\right )^{3/2}}{6 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 {\sqrt {a+b x^3} \left (A+B x^3\right )}{x^7} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx,x,x^3\right )\\ &=-\frac {A \left (a+b x^3\right )^{3/2}}{6 a x^6}+\frac {\left (-\frac {A b}{2}+2 a B\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,x^3\right )}{6 a}\\ &=\frac {(A b-4 a B) \sqrt {a+b x^3}}{12 a x^3}-\frac {A \left (a+b x^3\right )^{3/2}}{6 a x^6}-\frac {(b (A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{24 a}\\ &=\frac {(A b-4 a B) \sqrt {a+b x^3}}{12 a x^3}-\frac {A \left (a+b x^3\right )^{3/2}}{6 a x^6}-\frac {(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 )}{12 a}\\ &=\frac {(A b-4 a B) \sqrt {a+b x^3}}{12 a x^3}-\frac {A \left (a+b x^3\right )^{3/2}}{6 a x^6}+\frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 93, normalized size = 1.06 \begin {gather*} \frac {-\left (a+b x^3\right ) \left (2 a \left (A+2 B x^3\right )+A b x^3\right )-b x^6 \sqrt {\frac {b x^3}{a}+1} (4 a B-A b) \tanh ^{-1}\left (\sqrt {\frac {b x^3}{a}+1}\right )}{12 a x^6 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*x^3)*(A*b*x^3 + 2*a*(A + 2*B*x^3))) - b*(-(A*b) + 4*a*B)*x^6*Sqrt[1 + (b*x^3)/a]*ArcTanh[Sqrt[1 + (b
*x^3)/a]])/(12*a*x^6*Sqrt[a + b*x^3])

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IntegrateAlgebraic [A]  time = 0.12, size = 79, normalized size = 0.90 \begin {gather*} \frac {\left (A b^2-4 a b B\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}}+\frac {\sqrt {a+b x^3} \left (-2 a A-4 a B x^3-A b x^3\right )}{12 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.76, size = 172, normalized size = 1.95 \begin {gather*} \left [-\frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {a} x^{6} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left ({\left (4 \, B a^{2} + A a b\right )} x^{3} + 2 \, A a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a^{2} x^{6}}, \frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) - {\left ({\left (4 \, B a^{2} + A a b\right )} x^{3} + 2 \, A a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a^{2} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^7,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 120, normalized size = 1.36 \begin {gather*} \frac {\frac {{\left (4 \, B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x^{3} + a} B a^{2} b^{2} + {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b^{3} + \sqrt {b x^{3} + a} A a b^{3}}{a b^{2} x^{6}}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^7,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 96, normalized size = 1.09 \begin {gather*} \left (\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}\, b}{12 a \,x^{3}}-\frac {\sqrt {b \,x^{3}+a}}{6 x^{6}}\right ) A +\left (-\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \sqrt {a}}-\frac {\sqrt {b \,x^{3}+a}}{3 x^{3}}\right ) B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)*(b*x^3+a)^(1/2)/x^7,x)

[Out]

B*(-1/3*b*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(1/2)-1/3*(b*x^3+a)^(1/2)/x^3)+A*(1/12*b^2*arctanh((b*x^3+a)^(1/2
)/a^(1/2))/a^(3/2)-1/6*(b*x^3+a)^(1/2)/x^6-1/12*(b*x^3+a)^(1/2)/a*b/x^3)

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maxima [B]  time = 1.32, size = 158, normalized size = 1.80 \begin {gather*} -\frac {1}{24} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b x^{3} + a} a b^{2}\right )}}{{\left (b x^{3} + a\right )}^{2} a - 2 \, {\left (b x^{3} + a\right )} a^{2} + a^{3}}\right )} A + \frac {1}{6} \, {\left (\frac {b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 \, \sqrt {b x^{3} + a}}{x^{3}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^7,x, algorithm="maxima")

[Out]

-1/24*(b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(3/2) + 2*((b*x^3 + a)^(3/2)*b^2 + s
qrt(b*x^3 + a)*a*b^2)/((b*x^3 + a)^2*a - 2*(b*x^3 + a)*a^2 + a^3))*A + 1/6*(b*log((sqrt(b*x^3 + a) - sqrt(a))/
(sqrt(b*x^3 + a) + sqrt(a)))/sqrt(a) - 2*sqrt(b*x^3 + a)/x^3)*B

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mupad [B]  time = 3.12, size = 93, normalized size = 1.06 \begin {gather*} \frac {b\,\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 (A\,b-4\,B\,a\right )}{24\,a^{3/2}}-\frac {\left (4\,B\,a^2+A\,b\,a\right )\,\sqrt {b\,x^3+a}}{12\,a^2\,x^3}-\frac {A\,\sqrt {b\,x^3+a}}{6\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^(1/2))/x^7,x)

[Out]

(b*log((((a + b*x^3)^(1/2) - a^(1/2))*((a + b*x^3)^(1/2) + a^(1/2))^3)/x^6)*(A*b - 4*B*a))/(24*a^(3/2)) - ((4*
B*a^2 + A*a*b)*(a + b*x^3)^(1/2))/(12*a^2*x^3) - (A*(a + b*x^3)^(1/2))/(6*x^6)

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sympy [B]  time = 129.14, size = 160, normalized size = 1.82 \begin {gather*} - \frac {A a}{6 \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {A \sqrt {b}}{4 x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {A b^{\frac {3}{2}}}{12 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{12 a^{\frac {3}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 x^{\frac {3}{2}}} - \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)*(b*x**3+a)**(1/2)/x**7,x)

[Out]

-A*a/(6*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) - A*sqrt(b)/(4*x**(9/2)*sqrt(a/(b*x**3) + 1)) - A*b**(3/2)/(12
*a*x**(3/2)*sqrt(a/(b*x**3) + 1)) + A*b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(12*a**(3/2)) - B*sqrt(b)*sqrt(a/
(b*x**3) + 1)/(3*x**(3/2)) - B*b*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(3*sqrt(a))

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