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

Optimal. Leaf size=120 \[ \frac {b^2 (A b-6 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}+\frac {b \sqrt {a+b x^2} (A b-6 a B)}{16 a x^2}+\frac {\left (a+b x^2\right )^{3/2} (A b-6 a B)}{24 a x^4}-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6} \]

[Out]

1/24*(A*b-6*B*a)*(b*x^2+a)^(3/2)/a/x^4-1/6*A*(b*x^2+a)^(5/2)/a/x^6+1/16*b^2*(A*b-6*B*a)*arctanh((b*x^2+a)^(1/2
)/a^(1/2))/a^(3/2)+1/16*b*(A*b-6*B*a)*(b*x^2+a)^(1/2)/a/x^2

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Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, 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, 47, 63, 208} \[ \frac {b^2 (A b-6 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}+\frac {\left (a+b x^2\right )^{3/2} (A b-6 a B)}{24 a x^4}+\frac {b \sqrt {a+b x^2} (A b-6 a B)}{16 a x^2}-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A*b - 6*a*B)*Sqrt[a + b*x^2])/(16*a*x^2) + ((A*b - 6*a*B)*(a + b*x^2)^(3/2))/(24*a*x^4) - (A*(a + b*x^2)^(
5/2))/(6*a*x^6) + (b^2*(A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(16*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 {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2} (A+B x)}{x^4} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6}+\frac {\left (-\frac {A b}{2}+3 a B\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3} \, dx,x,x^2\right )}{6 a}\\ &=\frac {(A b-6 a B) \left (a+b x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6}-\frac {(b (A b-6 a B)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,x^2\right )}{16 a}\\ &=\frac {b (A b-6 a B) \sqrt {a+b x^2}}{16 a x^2}+\frac {(A b-6 a B) \left (a+b x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6}-\frac {\left (b^2 (A b-6 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{32 a}\\ &=\frac {b (A b-6 a B) \sqrt {a+b x^2}}{16 a x^2}+\frac {(A b-6 a B) \left (a+b x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6}-\frac {(b (A b-6 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{16 a}\\ &=\frac {b (A b-6 a B) \sqrt {a+b x^2}}{16 a x^2}+\frac {(A b-6 a B) \left (a+b x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{5/2}}{6 a x^6}+\frac {b^2 (A b-6 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 119, normalized size = 0.99 \[ \frac {-\left (a+b x^2\right ) \left (4 a^2 \left (2 A+3 B x^2\right )+2 a b x^2 \left (7 A+15 B x^2\right )+3 A b^2 x^4\right )-3 b^2 x^6 \sqrt {\frac {b x^2}{a}+1} (6 a B-A b) \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{48 a x^6 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*x^2)*(3*A*b^2*x^4 + 4*a^2*(2*A + 3*B*x^2) + 2*a*b*x^2*(7*A + 15*B*x^2))) - 3*b^2*(-(A*b) + 6*a*B)*x^
6*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/(48*a*x^6*Sqrt[a + b*x^2])

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fricas [A]  time = 0.66, size = 222, normalized size = 1.85 \[ \left [-\frac {3 \, {\left (6 \, B a b^{2} - A b^{3}\right )} \sqrt {a} x^{6} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (10 \, B a^{2} b + A a b^{2}\right )} x^{4} + 8 \, A a^{3} + 2 \, {\left (6 \, B a^{3} + 7 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, a^{2} x^{6}}, \frac {3 \, {\left (6 \, B a b^{2} - A b^{3}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, {\left (10 \, B a^{2} b + A a b^{2}\right )} x^{4} + 8 \, A a^{3} + 2 \, {\left (6 \, B a^{3} + 7 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a^{2} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(6*B*a*b^2 - A*b^3)*sqrt(a)*x^6*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*(10*B*a^2
*b + A*a*b^2)*x^4 + 8*A*a^3 + 2*(6*B*a^3 + 7*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^2*x^6), 1/48*(3*(6*B*a*b^2 - A*
b^3)*sqrt(-a)*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (3*(10*B*a^2*b + A*a*b^2)*x^4 + 8*A*a^3 + 2*(6*B*a^3 + 7*
A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^2*x^6)]

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giac [A]  time = 0.33, size = 159, normalized size = 1.32 \[ \frac {\frac {3 \, {\left (6 \, B a b^{3} - A b^{4}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {30 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a b^{3} - 48 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} b^{3} + 18 \, \sqrt {b x^{2} + a} B a^{3} b^{3} + 3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{4} + 8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{4} - 3 \, \sqrt {b x^{2} + a} A a^{2} b^{4}}{a b^{3} x^{6}}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(6*B*a*b^3 - A*b^4)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a) - (30*(b*x^2 + a)^(5/2)*B*a*b^3 - 48
*(b*x^2 + a)^(3/2)*B*a^2*b^3 + 18*sqrt(b*x^2 + a)*B*a^3*b^3 + 3*(b*x^2 + a)^(5/2)*A*b^4 + 8*(b*x^2 + a)^(3/2)*
A*a*b^4 - 3*sqrt(b*x^2 + a)*A*a^2*b^4)/(a*b^3*x^6))/b

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maple [B]  time = 0.01, size = 233, normalized size = 1.94 \[ \frac {A \,b^{3} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {3}{2}}}-\frac {3 B \,b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 \sqrt {a}}-\frac {\sqrt {b \,x^{2}+a}\, A \,b^{3}}{16 a^{2}}+\frac {3 \sqrt {b \,x^{2}+a}\, B \,b^{2}}{8 a}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{3}}{48 a^{3}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{2}}{8 a^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{2}}{48 a^{3} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B b}{8 a^{2} x^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A b}{24 a^{2} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B}{4 a \,x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*A*(b*x^2+a)^(5/2)/a/x^6+1/24*A*b/a^2/x^4*(b*x^2+a)^(5/2)+1/48*A*b^2/a^3/x^2*(b*x^2+a)^(5/2)-1/48*A*b^3/a^
3*(b*x^2+a)^(3/2)+1/16*A*b^3/a^(3/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-1/16*A*b^3/a^2*(b*x^2+a)^(1/2)-1/4*
B/a/x^4*(b*x^2+a)^(5/2)-1/8*B*b/a^2/x^2*(b*x^2+a)^(5/2)+1/8*B*b^2/a^2*(b*x^2+a)^(3/2)-3/8*B*b^2/a^(1/2)*ln((2*
a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+3/8*B*b^2/a*(b*x^2+a)^(1/2)

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maxima [B]  time = 1.09, size = 210, normalized size = 1.75 \[ -\frac {3 \, B b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {A b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B b^{2}}{8 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3}}{48 \, a^{3}} - \frac {\sqrt {b x^{2} + a} A b^{3}}{16 \, a^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{8 \, a^{2} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{48 \, a^{3} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{4 \, a x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{6 \, a x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*B*b^2*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 1/16*A*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) + 1/8*(b*x
^2 + a)^(3/2)*B*b^2/a^2 + 3/8*sqrt(b*x^2 + a)*B*b^2/a - 1/48*(b*x^2 + a)^(3/2)*A*b^3/a^3 - 1/16*sqrt(b*x^2 + a
)*A*b^3/a^2 - 1/8*(b*x^2 + a)^(5/2)*B*b/(a^2*x^2) + 1/48*(b*x^2 + a)^(5/2)*A*b^2/(a^3*x^2) - 1/4*(b*x^2 + a)^(
5/2)*B/(a*x^4) + 1/24*(b*x^2 + a)^(5/2)*A*b/(a^2*x^4) - 1/6*(b*x^2 + a)^(5/2)*A/(a*x^6)

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mupad [B]  time = 2.63, size = 130, normalized size = 1.08 \[ \frac {A\,a\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {5\,B\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4}-\frac {3\,B\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,\sqrt {a}}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}+\frac {3\,B\,a\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a\,x^6}-\frac {A\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a*(a + b*x^2)^(1/2))/(16*x^6) - (5*B*(a + b*x^2)^(3/2))/(8*x^4) - (A*b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2
))*1i)/(16*a^(3/2)) - (3*B*b^2*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(8*a^(1/2)) - (A*(a + b*x^2)^(3/2))/(6*x^6) +
(3*B*a*(a + b*x^2)^(1/2))/(8*x^4) - (A*(a + b*x^2)^(5/2))/(16*a*x^6)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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