3.5.20 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx\) [420]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.05, 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 = {79, 43, 44, 65, 214} \begin {gather*} -\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} \end {gather*}

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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], 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 ) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.34, size = 129, normalized size = 0.73 \begin {gather*} -\frac {\sqrt {a+b x} \left (-45 A b^4 x^4+30 a b^3 x^3 (A+5 B x)+96 a^4 (4 A+5 B x)+16 a^3 b x (63 A+85 B x)+4 a^2 b^2 x^2 (186 A+295 B x)\right )}{1920 a^2 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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1920*(Sqrt[a + b*x]*(-45*A*b^4*x^4 + 30*a*b^3*x^3*(A + 5*B*x) + 96*a^4*(4*A + 5*B*x) + 16*a^3*b*x*(63*A + 8
5*B*x) + 4*a^2*b^2*x^2*(186*A + 295*B*x)))/(a^2*x^5) + (b^4*(-3*A*b + 10*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/
(128*a^(5/2))

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 141, normalized size = 0.80

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-45 A \,b^{4} x^{4}+150 B a \,b^{3} x^{4}+30 A a \,b^{3} x^{3}+1180 B \,a^{2} b^{2} x^{3}+744 A \,a^{2} b^{2} x^{2}+1360 B \,a^{3} b \,x^{2}+1008 A \,a^{3} b x +480 B \,a^{4} x +384 A \,a^{4}\right )}{1920 x^{5} a^{2}}-\frac {b^{4} \left (3 A b -10 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {5}{2}}}\)	\(131\)
derivativedivides	\(2 b^{4} \left (-\frac {-\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\left (\frac {A b}{10}-\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {7 a \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {a^{2} \left (3 A b -10 B a \right ) \sqrt {b x +a}}{256}}{b^{5} x^{5}}-\frac {\left (3 A b -10 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}\right )\)	\(141\)
default	\(2 b^{4} \left (-\frac {-\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\left (\frac {A b}{10}-\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {7 a \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {a^{2} \left (3 A b -10 B a \right ) \sqrt {b x +a}}{256}}{b^{5} x^{5}}-\frac {\left (3 A b -10 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}\right )\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))/b^5/x^5-1/256*(3*A*b-10
*B*a)/a^(5/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 233, normalized size = 1.32 \begin {gather*} -\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 )} \end {gather*}

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))

________________________________________________________________________________________

Fricas [A]
time = 0.95, size = 307, normalized size = 1.73 \begin {gather*} \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 ] \end {gather*}

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)]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 2120 vs. \(2 (162) = 324\).
time = 107.77, size = 2120, normalized size = 11.98 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1930*A*a**7*b**5*sqrt(a + b*x)/(5120*a**10 + 6400*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**
3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)**5) + 4740*A*a**6*b**5*(a + b*x)**(3/2)/(5120*a**10 + 6400*a*
*9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)**5)
- 5376*A*a**5*b**5*(a + b*x)**(5/2)/(5120*a**10 + 6400*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*
x)**3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)**5) - 1674*A*a**5*b**5*sqrt(a + b*x)/(-1152*a**8 - 1536*a
**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) + 2940*A*a**4*b**5*(a + b*x
)**(7/2)/(5120*a**10 + 6400*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**3 - 6400*a**6*(a + b*x)
**4 + 1280*a**5*(a + b*x)**5) + 3066*A*a**4*b**5*(a + b*x)**(3/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a +
b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 630*A*a**3*b**5*(a + b*x)**(9/2)/(5120*a**10 + 64
00*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)
**5) - 2310*A*a**3*b**5*(a + b*x)**(5/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a +
b*x)**3 + 384*a**4*(a + b*x)**4) - 198*A*a**3*b**5*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)
**2 + 48*a**3*(a + b*x)**3) - 63*A*a**3*b**5*sqrt(a**(-11))*log(-a**6*sqrt(a**(-11)) + sqrt(a + b*x))/256 + 63
*A*a**3*b**5*sqrt(a**(-11))*log(a**6*sqrt(a**(-11)) + sqrt(a + b*x))/256 + 630*A*a**2*b**5*(a + b*x)**(7/2)/(-
1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) + 240*A*a
**2*b**5*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 105*A*a**2
*b**5*sqrt(a**(-9))*log(-a**5*sqrt(a**(-9)) + sqrt(a + b*x))/128 - 105*A*a**2*b**5*sqrt(a**(-9))*log(a**5*sqrt
(a**(-9)) + sqrt(a + b*x))/128 - 90*A*a*b**5*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2
+ 48*a**3*(a + b*x)**3) - 10*A*a*b**5*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) - 15*A*a*b**
5*sqrt(a**(-7))*log(-a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 15*A*a*b**5*sqrt(a**(-7))*log(a**4*sqrt(a**(-7))
+ sqrt(a + b*x))/16 + 6*A*b**5*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*A*b**5*sqrt
(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*A*b**5*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a
+ b*x))/8 - 558*B*a**6*b**4*sqrt(a + b*x)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a
+ b*x)**3 + 384*a**4*(a + b*x)**4) + 1022*B*a**5*b**4*(a + b*x)**(3/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6
*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 770*B*a**4*b**4*(a + b*x)**(5/2)/(-1152*a**8
- 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 198*B*a**4*b**4*
sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 210*B*a**3*b**4*(a + b
*x)**(7/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)*
*4) + 240*B*a**3*b**4*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3)
+ 35*B*a**3*b**4*sqrt(a**(-9))*log(-a**5*sqrt(a**(-9)) + sqrt(a + b*x))/128 - 35*B*a**3*b**4*sqrt(a**(-9))*lo
g(a**5*sqrt(a**(-9)) + sqrt(a + b*x))/128 - 90*B*a**2*b**4*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4
*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 30*B*a**2*b**4*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)
**2) - 15*B*a**2*b**4*sqrt(a**(-7))*log(-a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 15*B*a**2*b**4*sqrt(a**(-7))
*log(a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 18*B*a*b**4*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a
+ b*x)**2) + 9*B*a*b**4*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 9*B*a*b**4*sqrt(a**(-5))*lo
g(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - B*b**4*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + B*
b**4*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 - B*b**3*sqrt(a + b*x)/(a*x)

________________________________________________________________________________________

Giac [A]
time = 1.05, size = 208, normalized size = 1.18 \begin {gather*} -\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} \end {gather*}

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

________________________________________________________________________________________

Mupad [B]
time = 0.13, size = 217, normalized size = 1.23 \begin {gather*} \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}} \end {gather*}

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))

________________________________________________________________________________________