3.3.18 \(\int \frac {A+B x^3}{x^7 \sqrt {a+b x^3}} \, dx\) [218]

Optimal. Leaf size=90 \[ -\frac {A \sqrt {a+b x^3}}{6 a x^6}+\frac {(3 A b-4 a B) \sqrt {a+b x^3}}{12 a^2 x^3}-\frac {b (3 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 90, 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 = {457, 79, 44, 65, 214} \begin {gather*} -\frac {b (3 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{5/2}}+\frac {\sqrt {a+b x^3} (3 A b-4 a B)}{12 a^2 x^3}-\frac {A \sqrt {a+b x^3}}{6 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(A*Sqrt[a + b*x^3])/(a*x^6) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^3])/(12*a^2*x^3) - (b*(3*A*b - 4*a*B)*ArcTanh
[Sqrt[a + b*x^3]/Sqrt[a]])/(12*a^(5/2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

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^7 \sqrt {a+b x^3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=-\frac {A \sqrt {a+b x^3}}{6 a x^6}+\frac {\left (-\frac {3 A b}{2}+2 a B\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^3\right )}{6 a}\\ &=-\frac {A \sqrt {a+b x^3}}{6 a x^6}+\frac {(3 A b-4 a B) \sqrt {a+b x^3}}{12 a^2 x^3}+\frac {(b (3 A b-4 a B)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{24 a^2}\\ &=-\frac {A \sqrt {a+b x^3}}{6 a x^6}+\frac {(3 A b-4 a B) \sqrt {a+b x^3}}{12 a^2 x^3}+\frac {(3 A b-4 a B) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{12 a^2}\\ &=-\frac {A \sqrt {a+b x^3}}{6 a x^6}+\frac {(3 A b-4 a B) \sqrt {a+b x^3}}{12 a^2 x^3}-\frac {b (3 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 78, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a+b x^3} \left (-2 a A+3 A b x^3-4 a B x^3\right )}{12 a^2 x^6}+\frac {b (-3 A b+4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.35, size = 102, normalized size = 1.13

method	result	size
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (-3 A b \,x^{3}+4 B a \,x^{3}+2 A a \right )}{12 a^{2} x^{6}}-\frac {b \left (3 A b -4 B a \right ) \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {5}{2}}}\)	\(67\)
elliptic	\(-\frac {b \left (3 A b -4 B a \right ) \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {5}{2}}}-\frac {A \sqrt {b \,x^{3}+a}}{6 x^{6} a}+\frac {\left (3 A b -4 B a \right ) \sqrt {b \,x^{3}+a}}{12 a^{2} x^{3}}\)	\(75\)
default	\(A \left (-\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {5}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{6 x^{6} a}+\frac {b \sqrt {b \,x^{3}+a}}{4 a^{2} x^{3}}\right )+B \left (\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{3 a \,x^{3}}\right )\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (74) = 148\).
time = 0.49, size = 178, normalized size = 1.98 \begin {gather*} -\frac {1}{6} \, B {\left (\frac {2 \, \sqrt {b x^{3} + a} b}{{\left (b x^{3} + a\right )} a - a^{2}} + \frac {b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}}\right )} + \frac {1}{24} \, A {\left (\frac {3 \, b^{2} \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 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b x^{3} + a} a b^{2}\right )}}{{\left (b x^{3} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{3} + a\right )} a^{3} + a^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*B*(2*sqrt(b*x^3 + a)*b/((b*x^3 + a)*a - a^2) + b*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(
a)))/a^(3/2)) + 1/24*A*(3*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2) + 2*(3*(b*x
^3 + a)^(3/2)*b^2 - 5*sqrt(b*x^3 + a)*a*b^2)/((b*x^3 + a)^2*a^2 - 2*(b*x^3 + a)*a^3 + a^4))

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Fricas [A]
time = 1.62, size = 173, normalized size = 1.92 \begin {gather*} \left [-\frac {{\left (4 \, B a b - 3 \, 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} - 3 \, A a b\right )} x^{3} + 2 \, A a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a^{3} x^{6}}, -\frac {{\left (4 \, B a b - 3 \, 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} - 3 \, A a b\right )} x^{3} + 2 \, A a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a^{3} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (80) = 160\).
time = 25.94, size = 163, normalized size = 1.81 \begin {gather*} - \frac {A}{6 \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {A \sqrt {b}}{12 a x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {A b^{\frac {3}{2}}}{4 a^{2} 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 )}}{4 a^{\frac {5}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 a x^{\frac {3}{2}}} + \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A/(6*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) + A*sqrt(b)/(12*a*x**(9/2)*sqrt(a/(b*x**3) + 1)) + A*b**(3/2)/(4
*a**2*x**(3/2)*sqrt(a/(b*x**3) + 1)) - A*b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(4*a**(5/2)) - B*sqrt(b)*sqrt(
a/(b*x**3) + 1)/(3*a*x**(3/2)) + B*b*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(3*a**(3/2))

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Giac [A]
time = 1.05, size = 121, normalized size = 1.34 \begin {gather*} -\frac {\frac {{\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \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} - 3 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b^{3} + 5 \, \sqrt {b x^{3} + a} A a b^{3}}{a^{2} b^{2} x^{6}}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.99, size = 95, normalized size = 1.06 \begin {gather*} \frac {\sqrt {b\,x^3+a}\,\left (3\,A\,b-4\,B\,a\right )}{12\,a^2\,x^3}-\frac {A\,\sqrt {b\,x^3+a}}{6\,a\,x^6}+\frac {b\,\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-4\,B\,a\right )}{24\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + b*x^3)^(1/2)*(3*A*b - 4*B*a))/(12*a^2*x^3) - (A*(a + b*x^3)^(1/2))/(6*a*x^6) + (b*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 - 4*B*a))/(24*a^(5/2))

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