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

Optimal. Leaf size=115 \[ -\frac{\left (a+b x^3\right )^{3/2} (4 a B+A b)}{12 a x^3}+\frac{b \sqrt{a+b x^3} (4 a B+A b)}{4 a}-\frac{b (4 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0868472, antiderivative size = 115, normalized size of antiderivative = 1., 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 = {446, 78, 47, 50, 63, 208} \[ -\frac{\left (a+b x^3\right )^{3/2} (4 a B+A b)}{12 a x^3}+\frac{b \sqrt{a+b x^3} (4 a B+A b)}{4 a}-\frac{b (4 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 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{\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^7} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x^3} \, dx,x,x^3\right )\\ &=-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6}+\frac{(A b+4 a B) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,x^3\right )}{12 a}\\ &=-\frac{(A b+4 a B) \left (a+b x^3\right )^{3/2}}{12 a x^3}-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6}+\frac{(b (A b+4 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^3\right )}{8 a}\\ &=\frac{b (A b+4 a B) \sqrt{a+b x^3}}{4 a}-\frac{(A b+4 a B) \left (a+b x^3\right )^{3/2}}{12 a x^3}-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6}+\frac{1}{8} (b (A b+4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )\\ &=\frac{b (A b+4 a B) \sqrt{a+b x^3}}{4 a}-\frac{(A b+4 a B) \left (a+b x^3\right )^{3/2}}{12 a x^3}-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6}+\frac{1}{4} (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 )\\ &=\frac{b (A b+4 a B) \sqrt{a+b x^3}}{4 a}-\frac{(A b+4 a B) \left (a+b x^3\right )^{3/2}}{12 a x^3}-\frac{A \left (a+b x^3\right )^{5/2}}{6 a x^6}-\frac{b (A b+4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 \sqrt{a}}\\ \end{align*}

Mathematica [C]  time = 0.026296, size = 59, normalized size = 0.51 \[ \frac{\left (a+b x^3\right )^{5/2} \left (b x^6 (4 a B+A b) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{b x^3}{a}+1\right )-5 a^2 A\right )}{30 a^3 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(5/2)*(-5*a^2*A + b*(A*b + 4*a*B)*x^6*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x^3)/a]))/(30*a^3*x^6
)

________________________________________________________________________________________

Maple [A]  time = 0.023, size = 107, normalized size = 0.9 \begin{align*} A \left ( -{\frac{a}{6\,{x}^{6}}\sqrt{b{x}^{3}+a}}-{\frac{5\,b}{12\,{x}^{3}}\sqrt{b{x}^{3}+a}}-{\frac{{b}^{2}}{4}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a}}}} \right ) +B \left ( -{\frac{a}{3\,{x}^{3}}\sqrt{b{x}^{3}+a}}+{\frac{2\,b}{3}\sqrt{b{x}^{3}+a}}-\sqrt{a}b{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.81928, size = 439, normalized size = 3.82 \begin{align*} \left [\frac{3 \,{\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 (8 \, B a b x^{6} -{\left (4 \, B a^{2} + 5 \, A a b\right )} x^{3} - 2 \, A a^{2}\right )} \sqrt{b x^{3} + a}}{24 \, a x^{6}}, \frac{3 \,{\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 (8 \, B a b x^{6} -{\left (4 \, B a^{2} + 5 \, A a b\right )} x^{3} - 2 \, A a^{2}\right )} \sqrt{b x^{3} + a}}{12 \, a x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*(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*(8*B*a*b*x^6 - (
4*B*a^2 + 5*A*a*b)*x^3 - 2*A*a^2)*sqrt(b*x^3 + a))/(a*x^6), 1/12*(3*(4*B*a*b + A*b^2)*sqrt(-a)*x^6*arctan(sqrt
(b*x^3 + a)*sqrt(-a)/a) + (8*B*a*b*x^6 - (4*B*a^2 + 5*A*a*b)*x^3 - 2*A*a^2)*sqrt(b*x^3 + a))/(a*x^6)]

________________________________________________________________________________________

Sympy [B]  time = 62.9708, size = 243, normalized size = 2.11 \begin{align*} - \frac{A a^{2}}{6 \sqrt{b} x^{\frac{15}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} - \frac{A a \sqrt{b}}{4 x^{\frac{9}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} - \frac{A b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}}{3 x^{\frac{3}{2}}} - \frac{A b^{\frac{3}{2}}}{12 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 \sqrt{a}} - B \sqrt{a} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )} - \frac{B a \sqrt{b} \sqrt{\frac{a}{b x^{3}} + 1}}{3 x^{\frac{3}{2}}} + \frac{2 B a \sqrt{b}}{3 x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{2 B b^{\frac{3}{2}} x^{\frac{3}{2}}}{3 \sqrt{\frac{a}{b x^{3}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.15565, size = 177, normalized size = 1.54 \begin{align*} \frac{8 \, \sqrt{b x^{3} + a} B b^{2} + \frac{3 \,{\left (4 \, B a b^{2} + A b^{3}\right )} \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-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} + 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} A b^{3} - 3 \, \sqrt{b x^{3} + a} A a b^{3}}{b^{2} x^{6}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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