3.550 \(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^7} \, dx\)

Optimal. Leaf size=149 \[ \frac {5 b^2 \sqrt {a+b x^2} (6 a B+A b)}{16 a}-\frac {5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {5 b \left (a+b x^2\right )^{3/2} (6 a B+A b)}{48 a x^2}-\frac {\left (a+b x^2\right )^{5/2} (6 a B+A b)}{24 a x^4}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6} \]

[Out]

-5/48*b*(A*b+6*B*a)*(b*x^2+a)^(3/2)/a/x^2-1/24*(A*b+6*B*a)*(b*x^2+a)^(5/2)/a/x^4-1/6*A*(b*x^2+a)^(7/2)/a/x^6-5
/16*b^2*(A*b+6*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(1/2)+5/16*b^2*(A*b+6*B*a)*(b*x^2+a)^(1/2)/a

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Rubi [A]  time = 0.11, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {5 b^2 \sqrt {a+b x^2} (6 a B+A b)}{16 a}-\frac {5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {\left (a+b x^2\right )^{5/2} (6 a B+A b)}{24 a x^4}-\frac {5 b \left (a+b x^2\right )^{3/2} (6 a B+A b)}{48 a x^2}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^7,x]

[Out]

(5*b^2*(A*b + 6*a*B)*Sqrt[a + b*x^2])/(16*a) - (5*b*(A*b + 6*a*B)*(a + b*x^2)^(3/2))/(48*a*x^2) - ((A*b + 6*a*
B)*(a + b*x^2)^(5/2))/(24*a*x^4) - (A*(a + b*x^2)^(7/2))/(6*a*x^6) - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x
^2]/Sqrt[a]])/(16*Sqrt[a])

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 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 )^{5/2} \left (A+B x^2\right )}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac {(A b+6 a B) \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^3} \, dx,x,x^2\right )}{12 a}\\ &=-\frac {(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac {(5 b (A b+6 a B)) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )}{48 a}\\ &=-\frac {5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac {(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac {\left (5 b^2 (A b+6 a B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )}{32 a}\\ &=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x^2}}{16 a}-\frac {5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac {(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac {1}{32} \left (5 b^2 (A b+6 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x^2}}{16 a}-\frac {5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac {(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6}+\frac {1}{16} (5 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 )\\ &=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x^2}}{16 a}-\frac {5 b (A b+6 a B) \left (a+b x^2\right )^{3/2}}{48 a x^2}-\frac {(A b+6 a B) \left (a+b x^2\right )^{5/2}}{24 a x^4}-\frac {A \left (a+b x^2\right )^{7/2}}{6 a x^6}-\frac {5 b^2 (A b+6 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.41 \[ -\frac {\left (a+b x^2\right )^{7/2} \left (7 a^3 A+b^2 x^6 (6 a B+A b) \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {b x^2}{a}+1\right )\right )}{42 a^4 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^7,x]

[Out]

-1/42*((a + b*x^2)^(7/2)*(7*a^3*A + b^2*(A*b + 6*a*B)*x^6*Hypergeometric2F1[3, 7/2, 9/2, 1 + (b*x^2)/a]))/(a^4
*x^6)

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fricas [A]  time = 0.71, size = 241, normalized size = 1.62 \[ \left [\frac {15 \, {\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt {a} x^{6} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (48 \, B a b^{2} x^{6} - 3 \, {\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{4} - 8 \, A a^{3} - 2 \, {\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, a x^{6}}, \frac {15 \, {\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt {-a} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, B a b^{2} x^{6} - 3 \, {\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{4} - 8 \, A a^{3} - 2 \, {\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x, algorithm="fricas")

[Out]

[1/96*(15*(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*(48*B*a*b^2*
x^6 - 3*(18*B*a^2*b + 11*A*a*b^2)*x^4 - 8*A*a^3 - 2*(6*B*a^3 + 13*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a*x^6), 1/48
*(15*(6*B*a*b^2 + A*b^3)*sqrt(-a)*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (48*B*a*b^2*x^6 - 3*(18*B*a^2*b + 11*
A*a*b^2)*x^4 - 8*A*a^3 - 2*(6*B*a^3 + 13*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a*x^6)]

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giac [A]  time = 0.36, size = 167, normalized size = 1.12 \[ \frac {48 \, \sqrt {b x^{2} + a} B b^{3} + \frac {15 \, {\left (6 \, B a b^{3} + A b^{4}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {54 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a b^{3} - 96 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} b^{3} + 42 \, \sqrt {b x^{2} + a} B a^{3} b^{3} + 33 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{4} - 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{4} + 15 \, \sqrt {b x^{2} + a} A a^{2} b^{4}}{b^{3} x^{6}}}{48 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x, algorithm="giac")

[Out]

1/48*(48*sqrt(b*x^2 + a)*B*b^3 + 15*(6*B*a*b^3 + A*b^4)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (54*(b*x^2
+ a)^(5/2)*B*a*b^3 - 96*(b*x^2 + a)^(3/2)*B*a^2*b^3 + 42*sqrt(b*x^2 + a)*B*a^3*b^3 + 33*(b*x^2 + a)^(5/2)*A*b
^4 - 40*(b*x^2 + a)^(3/2)*A*a*b^4 + 15*sqrt(b*x^2 + a)*A*a^2*b^4)/(b^3*x^6))/b

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maple [B]  time = 0.02, size = 266, normalized size = 1.79 \[ -\frac {5 A \,b^{3} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 \sqrt {a}}-\frac {15 B \sqrt {a}\, b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8}+\frac {5 \sqrt {b \,x^{2}+a}\, A \,b^{3}}{16 a}+\frac {15 \sqrt {b \,x^{2}+a}\, B \,b^{2}}{8}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{3}}{48 a^{2}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{2}}{8 a}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{3}}{16 a^{3}}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,b^{2}}{8 a^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A \,b^{2}}{16 a^{3} x^{2}}-\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B b}{8 a^{2} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A b}{24 a^{2} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{4 a \,x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{6 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x)

[Out]

-1/6*A*(b*x^2+a)^(7/2)/a/x^6-1/24*A*b/a^2/x^4*(b*x^2+a)^(7/2)-1/16*A*b^2/a^3/x^2*(b*x^2+a)^(7/2)+1/16*A*b^3/a^
3*(b*x^2+a)^(5/2)+5/48*A*b^3/a^2*(b*x^2+a)^(3/2)-5/16*A*b^3/a^(1/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+5/16
*A*b^3/a*(b*x^2+a)^(1/2)-1/4*B/a/x^4*(b*x^2+a)^(7/2)-3/8*B*b/a^2/x^2*(b*x^2+a)^(7/2)+3/8*B*b^2/a^2*(b*x^2+a)^(
5/2)+5/8*B*b^2/a*(b*x^2+a)^(3/2)-15/8*B*b^2*a^(1/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+15/8*B*b^2*(b*x^2+a)
^(1/2)

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maxima [A]  time = 1.20, size = 243, normalized size = 1.63 \[ -\frac {15}{8} \, B \sqrt {a} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) - \frac {5 \, A b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, \sqrt {a}} + \frac {15}{8} \, \sqrt {b x^{2} + a} B b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{8 \, a^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{8 \, a} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{3}}{16 \, a^{3}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3}}{48 \, a^{2}} + \frac {5 \, \sqrt {b x^{2} + a} A b^{3}}{16 \, a} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{2}}{16 \, a^{3} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{4 \, a x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{6 \, a x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^7,x, algorithm="maxima")

[Out]

-15/8*B*sqrt(a)*b^2*arcsinh(a/(sqrt(a*b)*abs(x))) - 5/16*A*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 15/8*sq
rt(b*x^2 + a)*B*b^2 + 3/8*(b*x^2 + a)^(5/2)*B*b^2/a^2 + 5/8*(b*x^2 + a)^(3/2)*B*b^2/a + 1/16*(b*x^2 + a)^(5/2)
*A*b^3/a^3 + 5/48*(b*x^2 + a)^(3/2)*A*b^3/a^2 + 5/16*sqrt(b*x^2 + a)*A*b^3/a - 3/8*(b*x^2 + a)^(7/2)*B*b/(a^2*
x^2) - 1/16*(b*x^2 + a)^(7/2)*A*b^2/(a^3*x^2) - 1/4*(b*x^2 + a)^(7/2)*B/(a*x^4) - 1/24*(b*x^2 + a)^(7/2)*A*b/(
a^2*x^4) - 1/6*(b*x^2 + a)^(7/2)*A/(a*x^6)

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mupad [B]  time = 3.37, size = 150, normalized size = 1.01 \[ B\,b^2\,\sqrt {b\,x^2+a}-\frac {11\,A\,{\left (b\,x^2+a\right )}^{5/2}}{16\,x^6}+\frac {5\,A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {9\,B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4}-\frac {5\,A\,a^2\,\sqrt {b\,x^2+a}}{16\,x^6}+\frac {7\,B\,a^2\,\sqrt {b\,x^2+a}}{8\,x^4}+\frac {A\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,\sqrt {a}}+\frac {B\,\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^7,x)

[Out]

B*b^2*(a + b*x^2)^(1/2) - (11*A*(a + b*x^2)^(5/2))/(16*x^6) + (A*b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/
(16*a^(1/2)) + (B*a^(1/2)*b^2*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*15i)/8 + (5*A*a*(a + b*x^2)^(3/2))/(6*x^6)
- (9*B*a*(a + b*x^2)^(3/2))/(8*x^4) - (5*A*a^2*(a + b*x^2)^(1/2))/(16*x^6) + (7*B*a^2*(a + b*x^2)^(1/2))/(8*x^
4)

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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)**(5/2)*(B*x**2+A)/x**7,x)

[Out]

Timed out

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