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

Optimal. Leaf size=142 \[ \frac {5 b^3 (A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}+\frac {5 b^2 \sqrt {a+b x} (A b-8 a B)}{64 a x}+\frac {(a+b x)^{5/2} (A b-8 a B)}{24 a x^3}+\frac {5 b (a+b x)^{3/2} (A b-8 a B)}{96 a x^2}-\frac {A (a+b x)^{7/2}}{4 a x^4} \]

[Out]

5/96*b*(A*b-8*B*a)*(b*x+a)^(3/2)/a/x^2+1/24*(A*b-8*B*a)*(b*x+a)^(5/2)/x^3/a-1/4*A*(b*x+a)^(7/2)/a/x^4+5/64*b^3
*(A*b-8*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)+5/64*b^2*(A*b-8*B*a)*(b*x+a)^(1/2)/a/x

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Rubi [A]  time = 0.07, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 47, 63, 208} \[ \frac {5 b^3 (A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}+\frac {5 b^2 \sqrt {a+b x} (A b-8 a B)}{64 a x}+\frac {(a+b x)^{5/2} (A b-8 a B)}{24 a x^3}+\frac {5 b (a+b x)^{3/2} (A b-8 a B)}{96 a x^2}-\frac {A (a+b x)^{7/2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^2*(A*b - 8*a*B)*Sqrt[a + b*x])/(64*a*x) + (5*b*(A*b - 8*a*B)*(a + b*x)^(3/2))/(96*a*x^2) + ((A*b - 8*a*B)
*(a + b*x)^(5/2))/(24*a*x^3) - (A*(a + b*x)^(7/2))/(4*a*x^4) + (5*b^3*(A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt
[a]])/(64*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]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{x^5} \, dx &=-\frac {A (a+b x)^{7/2}}{4 a x^4}+\frac {\left (-\frac {A b}{2}+4 a B\right ) \int \frac {(a+b x)^{5/2}}{x^4} \, dx}{4 a}\\ &=\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {(5 b (A b-8 a B)) \int \frac {(a+b x)^{3/2}}{x^3} \, dx}{48 a}\\ &=\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {\left (5 b^2 (A b-8 a B)\right ) \int \frac {\sqrt {a+b x}}{x^2} \, dx}{64 a}\\ &=\frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {\left (5 b^3 (A b-8 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a}\\ &=\frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}-\frac {\left (5 b^2 (A b-8 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a}\\ &=\frac {5 b^2 (A b-8 a B) \sqrt {a+b x}}{64 a x}+\frac {5 b (A b-8 a B) (a+b x)^{3/2}}{96 a x^2}+\frac {(A b-8 a B) (a+b x)^{5/2}}{24 a x^3}-\frac {A (a+b x)^{7/2}}{4 a x^4}+\frac {5 b^3 (A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 124, normalized size = 0.87 \[ \frac {-(a+b x) \left (16 a^3 (3 A+4 B x)+8 a^2 b x (17 A+26 B x)+2 a b^2 x^2 (59 A+132 B x)+15 A b^3 x^3\right )-15 b^3 x^4 \sqrt {\frac {b x}{a}+1} (8 a B-A b) \tanh ^{-1}\left (\sqrt {\frac {b x}{a}+1}\right )}{192 a x^4 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*x)*(15*A*b^3*x^3 + 16*a^3*(3*A + 4*B*x) + 8*a^2*b*x*(17*A + 26*B*x) + 2*a*b^2*x^2*(59*A + 132*B*x)))
- 15*b^3*(-(A*b) + 8*a*B)*x^4*Sqrt[1 + (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]])/(192*a*x^4*Sqrt[a + b*x])

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fricas [A]  time = 0.60, size = 260, normalized size = 1.83 \[ \left [-\frac {15 \, {\left (8 \, B a b^{3} - A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} + 3 \, {\left (88 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{3} + 2 \, {\left (104 \, B a^{3} b + 59 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 17 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, \frac {15 \, {\left (8 \, B a b^{3} - A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (48 \, A a^{4} + 3 \, {\left (88 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{3} + 2 \, {\left (104 \, B a^{3} b + 59 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 17 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*(8*B*a*b^3 - A*b^4)*sqrt(a)*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^4 + 3*(88
*B*a^2*b^2 + 5*A*a*b^3)*x^3 + 2*(104*B*a^3*b + 59*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 17*A*a^3*b)*x)*sqrt(b*x + a))/
(a^2*x^4), 1/192*(15*(8*B*a*b^3 - A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (48*A*a^4 + 3*(88*B*a
^2*b^2 + 5*A*a*b^3)*x^3 + 2*(104*B*a^3*b + 59*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 17*A*a^3*b)*x)*sqrt(b*x + a))/(a^2
*x^4)]

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giac [A]  time = 1.62, size = 177, normalized size = 1.25 \[ \frac {\frac {15 \, {\left (8 \, B a b^{4} - A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {264 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} - 584 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 440 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} - 120 \, \sqrt {b x + a} B a^{4} b^{4} + 15 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} + 15 \, \sqrt {b x + a} A a^{3} b^{5}}{a b^{4} x^{4}}}{192 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(15*(8*B*a*b^4 - A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) - (264*(b*x + a)^(7/2)*B*a*b^4 - 584
*(b*x + a)^(5/2)*B*a^2*b^4 + 440*(b*x + a)^(3/2)*B*a^3*b^4 - 120*sqrt(b*x + a)*B*a^4*b^4 + 15*(b*x + a)^(7/2)*
A*b^5 + 73*(b*x + a)^(5/2)*A*a*b^5 - 55*(b*x + a)^(3/2)*A*a^2*b^5 + 15*sqrt(b*x + a)*A*a^3*b^5)/(a*b^4*x^4))/b

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maple [A]  time = 0.02, size = 118, normalized size = 0.83 \[ 2 \left (\frac {5 \left (A b -8 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}+\frac {\frac {55 \left (A b -8 B a \right ) \left (b x +a \right )^{\frac {3}{2}} a}{384}-\frac {\left (5 A b +88 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a}+\left (-\frac {73 A b}{384}+\frac {73 B a}{48}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {5}{128} A \,a^{2} b +\frac {5}{16} B \,a^{3}\right ) \sqrt {b x +a}}{b^{4} x^{4}}\right ) b^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*((-1/128*(5*A*b+88*B*a)/a*(b*x+a)^(7/2)+(73/48*B*a-73/384*A*b)*(b*x+a)^(5/2)+55/384*a*(A*b-8*B*a)*(b*x+a
)^(3/2)+(5/16*B*a^3-5/128*A*a^2*b)*(b*x+a)^(1/2))/x^4/b^4+5/128*(A*b-8*B*a)/a^(3/2)*arctanh((b*x+a)^(1/2)/a^(1
/2)))

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maxima [A]  time = 2.00, size = 194, normalized size = 1.37 \[ -\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (3 \, {\left (88 \, B a + 5 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 73 \, {\left (8 \, B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 55 \, {\left (8 \, B a^{3} - A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 15 \, {\left (8 \, B a^{4} - A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a b - 4 \, {\left (b x + a\right )}^{3} a^{2} b + 6 \, {\left (b x + a\right )}^{2} a^{3} b - 4 \, {\left (b x + a\right )} a^{4} b + a^{5} b} - \frac {15 \, {\left (8 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*b^4*(2*(3*(88*B*a + 5*A*b)*(b*x + a)^(7/2) - 73*(8*B*a^2 - A*a*b)*(b*x + a)^(5/2) + 55*(8*B*a^3 - A*a^2
*b)*(b*x + a)^(3/2) - 15*(8*B*a^4 - A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a*b - 4*(b*x + a)^3*a^2*b + 6*(b*x +
a)^2*a^3*b - 4*(b*x + a)*a^4*b + a^5*b) - 15*(8*B*a - A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt
(a)))/(a^(3/2)*b))

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mupad [B]  time = 0.45, size = 177, normalized size = 1.25 \[ \frac {5\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-8\,B\,a\right )}{64\,a^{3/2}}-\frac {\left (\frac {73\,A\,b^4}{192}-\frac {73\,B\,a\,b^3}{24}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {5\,A\,a^2\,b^4}{64}-\frac {5\,B\,a^3\,b^3}{8}\right )\,\sqrt {a+b\,x}+\left (\frac {55\,B\,a^2\,b^3}{24}-\frac {55\,A\,a\,b^4}{192}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {\left (5\,A\,b^4+88\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(A*b - 8*B*a))/(64*a^(3/2)) - (((73*A*b^4)/192 - (73*B*a*b^3)/24)*(a + b
*x)^(5/2) + ((5*A*a^2*b^4)/64 - (5*B*a^3*b^3)/8)*(a + b*x)^(1/2) + ((55*B*a^2*b^3)/24 - (55*A*a*b^4)/192)*(a +
b*x)^(3/2) + ((5*A*b^4 + 88*B*a*b^3)*(a + b*x)^(7/2))/(64*a))/((a + b*x)^4 - 4*a^3*(a + b*x) - 4*a*(a + b*x)^
3 + 6*a^2*(a + b*x)^2 + a^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+a)**(5/2)*(B*x+A)/x**5,x)

[Out]

Timed out

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