3.420 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx\)

Optimal. Leaf size=177 \[ -\frac {b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}}+\frac {b^3 \sqrt {a+b x} (3 A b-10 a B)}{128 a^2 x}+\frac {b^2 \sqrt {a+b x} (3 A b-10 a B)}{64 a x^2}+\frac {(a+b x)^{5/2} (3 A b-10 a B)}{40 a x^4}+\frac {b (a+b x)^{3/2} (3 A b-10 a B)}{48 a x^3}-\frac {A (a+b x)^{7/2}}{5 a x^5} \]

[Out]

1/48*b*(3*A*b-10*B*a)*(b*x+a)^(3/2)/x^3/a+1/40*(3*A*b-10*B*a)*(b*x+a)^(5/2)/a/x^4-1/5*A*(b*x+a)^(7/2)/a/x^5-1/
128*b^4*(3*A*b-10*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(5/2)+1/64*b^2*(3*A*b-10*B*a)*(b*x+a)^(1/2)/a/x^2+1/12
8*b^3*(3*A*b-10*B*a)*(b*x+a)^(1/2)/a^2/x

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Rubi [A]  time = 0.08, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {78, 47, 51, 63, 208} \[ \frac {b^3 \sqrt {a+b x} (3 A b-10 a B)}{128 a^2 x}-\frac {b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}}+\frac {b^2 \sqrt {a+b x} (3 A b-10 a B)}{64 a x^2}+\frac {b (a+b x)^{3/2} (3 A b-10 a B)}{48 a x^3}+\frac {(a+b x)^{5/2} (3 A b-10 a B)}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^6,x]

[Out]

(b^2*(3*A*b - 10*a*B)*Sqrt[a + b*x])/(64*a*x^2) + (b^3*(3*A*b - 10*a*B)*Sqrt[a + b*x])/(128*a^2*x) + (b*(3*A*b
- 10*a*B)*(a + b*x)^(3/2))/(48*a*x^3) + ((3*A*b - 10*a*B)*(a + b*x)^(5/2))/(40*a*x^4) - (A*(a + b*x)^(7/2))/(
5*a*x^5) - (b^4*(3*A*b - 10*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(5/2))

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 51

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 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]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx &=-\frac {A (a+b x)^{7/2}}{5 a x^5}+\frac {\left (-\frac {3 A b}{2}+5 a B\right ) \int \frac {(a+b x)^{5/2}}{x^5} \, dx}{5 a}\\ &=\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {(b (3 A b-10 a B)) \int \frac {(a+b x)^{3/2}}{x^4} \, dx}{16 a}\\ &=\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {\left (b^2 (3 A b-10 a B)\right ) \int \frac {\sqrt {a+b x}}{x^3} \, dx}{32 a}\\ &=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {\left (b^3 (3 A b-10 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{128 a}\\ &=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}+\frac {\left (b^4 (3 A b-10 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{256 a^2}\\ &=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}+\frac {\left (b^3 (3 A b-10 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{128 a^2}\\ &=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 58, normalized size = 0.33 \[ -\frac {(a+b x)^{7/2} \left (7 a^5 A+b^4 x^5 (10 a B-3 A b) \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {b x}{a}+1\right )\right )}{35 a^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^6,x]

[Out]

-1/35*((a + b*x)^(7/2)*(7*a^5*A + b^4*(-3*A*b + 10*a*B)*x^5*Hypergeometric2F1[7/2, 5, 9/2, 1 + (b*x)/a]))/(a^6
*x^5)

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fricas [A]  time = 0.62, size = 307, normalized size = 1.73 \[ \left [-\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {a} x^{5} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (384 \, A a^{5} + 15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{3840 \, a^{3} x^{5}}, -\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (384 \, A a^{5} + 15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{1920 \, a^{3} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^6,x, algorithm="fricas")

[Out]

[-1/3840*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(a)*x^5*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(384*A*a^5 +
15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^3)*x^3 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^2
+ 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(b*x + a))/(a^3*x^5), -1/1920*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(-a)*x^5*arct
an(sqrt(b*x + a)*sqrt(-a)/a) + (384*A*a^5 + 15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^
3)*x^3 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^2 + 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(b*x + a))/(a^3*x^5)]

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giac [A]  time = 1.78, size = 208, normalized size = 1.18 \[ -\frac {\frac {15 \, {\left (10 \, B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {150 \, {\left (b x + a\right )}^{\frac {9}{2}} B a b^{5} + 580 \, {\left (b x + a\right )}^{\frac {7}{2}} B a^{2} b^{5} - 1280 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{3} b^{5} + 700 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 150 \, \sqrt {b x + a} B a^{5} b^{5} - 45 \, {\left (b x + a\right )}^{\frac {9}{2}} A b^{6} + 210 \, {\left (b x + a\right )}^{\frac {7}{2}} A a b^{6} + 384 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 210 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 45 \, \sqrt {b x + a} A a^{4} b^{6}}{a^{2} b^{5} x^{5}}}{1920 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^6,x, algorithm="giac")

[Out]

-1/1920*(15*(10*B*a*b^5 - 3*A*b^6)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (150*(b*x + a)^(9/2)*B*a*b^
5 + 580*(b*x + a)^(7/2)*B*a^2*b^5 - 1280*(b*x + a)^(5/2)*B*a^3*b^5 + 700*(b*x + a)^(3/2)*B*a^4*b^5 - 150*sqrt(
b*x + a)*B*a^5*b^5 - 45*(b*x + a)^(9/2)*A*b^6 + 210*(b*x + a)^(7/2)*A*a*b^6 + 384*(b*x + a)^(5/2)*A*a^2*b^6 -
210*(b*x + a)^(3/2)*A*a^3*b^6 + 45*sqrt(b*x + a)*A*a^4*b^6)/(a^2*b^5*x^5))/b

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maple [A]  time = 0.02, size = 140, normalized size = 0.79 \[ 2 \left (-\frac {\left (3 A b -10 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}+\frac {-\frac {\left (3 A b -10 B a \right ) \sqrt {b x +a}\, a^{2}}{256}+\frac {7 \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}} a}{384}-\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\left (-\frac {A b}{10}+\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}}{b^{5} x^{5}}\right ) b^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^6,x)

[Out]

2*b^4*((1/256*(3*A*b-10*B*a)/a^2*(b*x+a)^(9/2)-1/384*(21*A*b+58*B*a)/a*(b*x+a)^(7/2)+(-1/10*A*b+1/3*B*a)*(b*x+
a)^(5/2)+7/384*a*(3*A*b-10*B*a)*(b*x+a)^(3/2)-1/256*a^2*(3*A*b-10*B*a)*(b*x+a)^(1/2))/x^5/b^5-1/256*(3*A*b-10*
B*a)/a^(5/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 2.10, size = 233, normalized size = 1.32 \[ -\frac {1}{3840} \, b^{5} {\left (\frac {2 \, {\left (15 \, {\left (10 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{\frac {9}{2}} + 10 \, {\left (58 \, B a^{2} + 21 \, A a b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 128 \, {\left (10 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 70 \, {\left (10 \, B a^{4} - 3 \, A a^{3} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 15 \, {\left (10 \, B a^{5} - 3 \, A a^{4} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{5} a^{2} b - 5 \, {\left (b x + a\right )}^{4} a^{3} b + 10 \, {\left (b x + a\right )}^{3} a^{4} b - 10 \, {\left (b x + a\right )}^{2} a^{5} b + 5 \, {\left (b x + a\right )} a^{6} b - a^{7} b} + \frac {15 \, {\left (10 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^6,x, algorithm="maxima")

[Out]

-1/3840*b^5*(2*(15*(10*B*a - 3*A*b)*(b*x + a)^(9/2) + 10*(58*B*a^2 + 21*A*a*b)*(b*x + a)^(7/2) - 128*(10*B*a^3
- 3*A*a^2*b)*(b*x + a)^(5/2) + 70*(10*B*a^4 - 3*A*a^3*b)*(b*x + a)^(3/2) - 15*(10*B*a^5 - 3*A*a^4*b)*sqrt(b*x
+ a))/((b*x + a)^5*a^2*b - 5*(b*x + a)^4*a^3*b + 10*(b*x + a)^3*a^4*b - 10*(b*x + a)^2*a^5*b + 5*(b*x + a)*a^
6*b - a^7*b) + 15*(10*B*a - 3*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(5/2)*b))

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mupad [B]  time = 0.13, size = 217, normalized size = 1.23 \[ \frac {\left (\frac {A\,b^5}{5}-\frac {2\,B\,a\,b^4}{3}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {3\,A\,a^2\,b^5}{128}-\frac {5\,B\,a^3\,b^4}{64}\right )\,\sqrt {a+b\,x}+\left (\frac {35\,B\,a^2\,b^4}{96}-\frac {7\,A\,a\,b^5}{64}\right )\,{\left (a+b\,x\right )}^{3/2}-\frac {\left (3\,A\,b^5-10\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{9/2}}{128\,a^2}+\frac {\left (21\,A\,b^5+58\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{7/2}}{192\,a}}{5\,a\,{\left (a+b\,x\right )}^4-5\,a^4\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^5-10\,a^2\,{\left (a+b\,x\right )}^3+10\,a^3\,{\left (a+b\,x\right )}^2+a^5}-\frac {b^4\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-10\,B\,a\right )}{128\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^6,x)

[Out]

(((A*b^5)/5 - (2*B*a*b^4)/3)*(a + b*x)^(5/2) + ((3*A*a^2*b^5)/128 - (5*B*a^3*b^4)/64)*(a + b*x)^(1/2) + ((35*B
*a^2*b^4)/96 - (7*A*a*b^5)/64)*(a + b*x)^(3/2) - ((3*A*b^5 - 10*B*a*b^4)*(a + b*x)^(9/2))/(128*a^2) + ((21*A*b
^5 + 58*B*a*b^4)*(a + b*x)^(7/2))/(192*a))/(5*a*(a + b*x)^4 - 5*a^4*(a + b*x) - (a + b*x)^5 - 10*a^2*(a + b*x)
^3 + 10*a^3*(a + b*x)^2 + a^5) - (b^4*atanh((a + b*x)^(1/2)/a^(1/2))*(3*A*b - 10*B*a))/(128*a^(5/2))

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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+a)**(5/2)*(B*x+A)/x**6,x)

[Out]

Timed out

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