3.3.33 \(\int \frac {A+B x^3}{x^7 (a+b x^3)^{3/2}} \, dx\) [233]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {457, 79, 44, 53, 65, 214} \begin {gather*} -\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 {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 {A}{6 a x^6 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[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))

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 53

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 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 \left (a+b x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {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 ) \text {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) \text {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)) \text {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) \text {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*}

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Mathematica [A]
time = 0.15, size = 100, normalized size = 0.85 \begin {gather*} \frac {-2 a^2 A+5 a A b x^3-4 a^2 B x^3+15 A b^2 x^6-12 a b B x^6}{12 a^3 x^6 \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*A + 5*a*A*b*x^3 - 4*a^2*B*x^3 + 15*A*b^2*x^6 - 12*a*b*B*x^6)/(12*a^3*x^6*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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Maple [A]
time = 0.38, size = 141, normalized size = 1.19

method	result	size
elliptic	\(-\frac {A \sqrt {b \,x^{3}+a}}{6 a^{2} x^{6}}+\frac {\left (7 A b -4 B a \right ) \sqrt {b \,x^{3}+a}}{12 a^{3} x^{3}}+\frac {2 b \left (A b -B a \right )}{3 a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {b \left (5 A b -4 B a \right ) \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\)	\(102\)
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (-7 A b \,x^{3}+4 B a \,x^{3}+2 A a \right )}{12 a^{3} x^{6}}+\frac {b \left (-\frac {2 \left (7 A b -4 B a \right )}{3 \sqrt {b \,x^{3}+a}}+3 a \left (5 A b -4 B a \right ) \left (\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\right )\right )}{8 a^{3}}\)	\(115\)
default	\(A \left (-\frac {\sqrt {b \,x^{3}+a}}{6 a^{2} x^{6}}+\frac {7 b \sqrt {b \,x^{3}+a}}{12 a^{3} x^{3}}+\frac {2 b^{2}}{3 a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {5 b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\right )+B \left (-\frac {2 b}{3 a^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{3}+a}}{3 a^{2} x^{3}}+\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )\)	\(141\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (98) = 196\).
time = 0.49, size = 215, normalized size = 1.82 \begin {gather*} \frac {1}{24} \, A {\left (\frac {2 \, {\left (15 \, {\left (b x^{3} + a\right )}^{2} b^{2} - 25 \, {\left (b x^{3} + a\right )} a b^{2} + 8 \, a^{2} b^{2}\right )}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x^{3} + a} a^{5}} + \frac {15 \, b^{2} \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}{6} \, B {\left (\frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )} b - 2 \, a b\right )}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{3} + a} a^{3}} + \frac {3 \, b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\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)^(3/2),x, algorithm="maxima")

[Out]

1/24*A*(2*(15*(b*x^3 + a)^2*b^2 - 25*(b*x^3 + a)*a*b^2 + 8*a^2*b^2)/((b*x^3 + a)^(5/2)*a^3 - 2*(b*x^3 + a)^(3/
2)*a^4 + sqrt(b*x^3 + a)*a^5) + 15*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(7/2)) -
1/6*B*(2*(3*(b*x^3 + a)*b - 2*a*b)/((b*x^3 + a)^(3/2)*a^2 - sqrt(b*x^3 + a)*a^3) + 3*b*log((sqrt(b*x^3 + a) -
sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2))

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Fricas [A]
time = 3.22, size = 289, normalized size = 2.45 \begin {gather*} \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 {gather*}

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 [A]
time = 50.66, size = 192, normalized size = 1.63 \begin {gather*} A \left (- \frac {1}{6 a \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {5 \sqrt {b}}{12 a^{2} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {5 b^{\frac {3}{2}}}{4 a^{3} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {5 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 a^{\frac {7}{2}}}\right ) + B \left (- \frac {1}{3 a \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {\sqrt {b}}{a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{a^{\frac {5}{2}}}\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)**(3/2),x)

[Out]

A*(-1/(6*a*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) + 5*sqrt(b)/(12*a**2*x**(9/2)*sqrt(a/(b*x**3) + 1)) + 5*b**
(3/2)/(4*a**3*x**(3/2)*sqrt(a/(b*x**3) + 1)) - 5*b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(4*a**(7/2))) + B*(-1/
(3*a*sqrt(b)*x**(9/2)*sqrt(a/(b*x**3) + 1)) - sqrt(b)/(a**2*x**(3/2)*sqrt(a/(b*x**3) + 1)) + b*asinh(sqrt(a)/(
sqrt(b)*x**(3/2)))/a**(5/2))

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Giac [A]
time = 0.69, size = 137, normalized size = 1.16 \begin {gather*} -\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 {gather*}

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)

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Mupad [B]
time = 3.18, size = 167, normalized size = 1.42 \begin {gather*} \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 (5\,A\,b-4\,B\,a\right )}{8\,a^{7/2}}-\frac {\left (4\,B\,a^2-7\,A\,a\,b\right )\,\sqrt {b\,x^3+a}}{12\,a^4\,x^3}-\frac {A\,\sqrt {b\,x^3+a}}{6\,a^2\,x^6}-\frac {\frac {a\,\left (\frac {7\,A\,b^3-4\,B\,a\,b^2}{12\,a^4}-\frac {5\,b^2\,\left (5\,A\,b-4\,B\,a\right )}{8\,a^4}\right )}{b}+\frac {3\,b\,\left (5\,A\,b-4\,B\,a\right )}{8\,a^3}}{\sqrt {b\,x^3+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*log((((a + b*x^3)^(1/2) - a^(1/2))^3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6)*(5*A*b - 4*B*a))/(8*a^(7/2)) - ((4
*B*a^2 - 7*A*a*b)*(a + b*x^3)^(1/2))/(12*a^4*x^3) - (A*(a + b*x^3)^(1/2))/(6*a^2*x^6) - ((a*((7*A*b^3 - 4*B*a*
b^2)/(12*a^4) - (5*b^2*(5*A*b - 4*B*a))/(8*a^4)))/b + (3*b*(5*A*b - 4*B*a))/(8*a^3))/(a + b*x^3)^(1/2)

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