\(\int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx\) [505]

Optimal result

Integrand size = 20, antiderivative size = 144 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {5 a^2 (6 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}} \]

[Out]

5/8*a^2*(6*A*b+B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(1/2)-2*A*(b*x+a)^(7/2)/a/x^(1/2)+5/12*(6*A*b+B*a
)*(b*x+a)^(3/2)*x^(1/2)+1/3*(6*A*b+B*a)*(b*x+a)^(5/2)*x^(1/2)/a+5/8*a*(6*A*b+B*a)*x^(1/2)*(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\frac {5 a^2 (a B+6 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}}+\frac {\sqrt {x} (a+b x)^{5/2} (a B+6 A b)}{3 a}+\frac {5}{12} \sqrt {x} (a+b x)^{3/2} (a B+6 A b)+\frac {5}{8} a \sqrt {x} \sqrt {a+b x} (a B+6 A b)-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}} \]

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(3/2),x]

[Out]

(5*a*(6*A*b + a*B)*Sqrt[x]*Sqrt[a + b*x])/8 + (5*(6*A*b + a*B)*Sqrt[x]*(a + b*x)^(3/2))/12 + ((6*A*b + a*B)*Sq
rt[x]*(a + b*x)^(5/2))/(3*a) - (2*A*(a + b*x)^(7/2))/(a*Sqrt[x]) + (5*a^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[
x])/Sqrt[a + b*x]])/(8*Sqrt[b])

Rule 52

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 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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {\left (2 \left (3 A b+\frac {a B}{2}\right )\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {x}} \, dx}{a} \\ & = \frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{6} (5 (6 A b+a B)) \int \frac {(a+b x)^{3/2}}{\sqrt {x}} \, dx \\ & = \frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{8} (5 a (6 A b+a B)) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx \\ & = \frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{16} \left (5 a^2 (6 A b+a B)\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx \\ & = \frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{8} \left (5 a^2 (6 A b+a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {1}{8} \left (5 a^2 (6 A b+a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right ) \\ & = \frac {5}{8} a (6 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {5}{12} (6 A b+a B) \sqrt {x} (a+b x)^{3/2}+\frac {(6 A b+a B) \sqrt {x} (a+b x)^{5/2}}{3 a}-\frac {2 A (a+b x)^{7/2}}{a \sqrt {x}}+\frac {5 a^2 (6 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (4 b^2 x^2 (3 A+2 B x)+2 a b x (27 A+13 B x)+a^2 (-48 A+33 B x)\right )}{24 \sqrt {x}}+\frac {5 a^2 (6 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 \sqrt {b}} \]

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(3/2),x]

[Out]

(Sqrt[a + b*x]*(4*b^2*x^2*(3*A + 2*B*x) + 2*a*b*x*(27*A + 13*B*x) + a^2*(-48*A + 33*B*x)))/(24*Sqrt[x]) + (5*a
^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(4*Sqrt[b])

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-8 b^{2} B \,x^{3}-12 A \,b^{2} x^{2}-26 B a b \,x^{2}-54 a A b x -33 a^{2} B x +48 a^{2} A \right )}{24 \sqrt {x}}+\frac {5 a^{2} \left (6 A b +B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 \sqrt {b}\, \sqrt {x}\, \sqrt {b x +a}}\)	\(119\)
default	\(\frac {\sqrt {b x +a}\, \left (16 B \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{3}+24 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}+52 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a \,x^{2}+90 A b \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} x +108 A a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} x +66 B \,a^{2} x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-96 A \,a^{2} \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{48 \sqrt {x}\, \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(202\)

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/24*(b*x+a)^(1/2)*(-8*B*b^2*x^3-12*A*b^2*x^2-26*B*a*b*x^2-54*A*a*b*x-33*B*a^2*x+48*A*a^2)/x^(1/2)+5/16*a^2*(
6*A*b+B*a)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))/b^(1/2)*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\left [\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (8 \, B b^{3} x^{3} - 48 \, A a^{2} b + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{2} + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b x}, -\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, B b^{3} x^{3} - 48 \, A a^{2} b + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{2} + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b x}\right ] \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(3/2),x, algorithm="fricas")

[Out]

[1/48*(15*(B*a^3 + 6*A*a^2*b)*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*B*b^3*x^3 - 48
*A*a^2*b + 2*(13*B*a*b^2 + 6*A*b^3)*x^2 + 3*(11*B*a^2*b + 18*A*a*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x), -1/24*(
15*(B*a^3 + 6*A*a^2*b)*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*B*b^3*x^3 - 48*A*a^2*b + 2*(
13*B*a*b^2 + 6*A*b^3)*x^2 + 3*(11*B*a^2*b + 18*A*a*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x)]

Sympy [A] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 478, normalized size of antiderivative = 3.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=- \frac {2 A a^{\frac {5}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + 2 A a^{\frac {3}{2}} b \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {2 A a^{\frac {3}{2}} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 4 A a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + 2 A b^{2} \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + B a^{\frac {5}{2}} \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {B a^{\frac {5}{2}} \sqrt {x}}{8 \sqrt {1 + \frac {b x}{a}}} - \frac {B a^{\frac {3}{2}} b x^{\frac {3}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} + \frac {5 B \sqrt {a} b^{2} x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} + \frac {9 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 \sqrt {b}} + 4 B a b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + \frac {B b^{3} x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**(3/2),x)

[Out]

-2*A*a**(5/2)/(sqrt(x)*sqrt(1 + b*x/a)) + 2*A*a**(3/2)*b*sqrt(x)*sqrt(1 + b*x/a) - 2*A*a**(3/2)*b*sqrt(x)/sqrt
(1 + b*x/a) + 4*A*a**2*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) + 2*A*b**2*Piecewise((-a**2*Piecewise((log(2*sqr
t(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(8*b) + a*sqrt(x
)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x**(3/2)/3, True)) + B*a**(5/2)*sqrt(x)*
sqrt(1 + b*x/a) - B*a**(5/2)*sqrt(x)/(8*sqrt(1 + b*x/a)) - B*a**(3/2)*b*x**(3/2)/(24*sqrt(1 + b*x/a)) + 5*B*sq
rt(a)*b**2*x**(5/2)/(12*sqrt(1 + b*x/a)) + 9*B*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b)) + 4*B*a*b*Piece
wise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sq
rt(b*x), True))/(8*b) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x**(3/2)
/3, True)) + B*b**3*x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/a))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (112) = 224\).

Time = 0.21 (sec) , antiderivative size = 487, normalized size of antiderivative = 3.38 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\frac {B b^{3} x^{4}}{3 \, \sqrt {b x^{2} + a x}} - \frac {7 \, B a b^{2} x^{3}}{12 \, \sqrt {b x^{2} + a x}} + \frac {35 \, B a^{2} b x^{2}}{24 \, \sqrt {b x^{2} + a x}} + \frac {51 \, B a^{3} x}{8 \, \sqrt {b x^{2} + a x}} + \frac {4 \, A a^{2} b x}{\sqrt {b x^{2} + a x}} - \frac {35 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, \sqrt {b}} - \frac {2 \, A a^{3}}{\sqrt {b x^{2} + a x}} + \frac {{\left (4 \, B a b^{3} + A b^{4}\right )} x^{3}}{2 \, \sqrt {b x^{2} + a x} b} - \frac {5 \, {\left (4 \, B a b^{3} + A b^{4}\right )} a x^{2}}{4 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{2}}{\sqrt {b x^{2} + a x} b} - \frac {15 \, {\left (4 \, B a b^{3} + A b^{4}\right )} a^{2} x}{4 \, \sqrt {b x^{2} + a x} b^{3}} + \frac {6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} a x}{\sqrt {b x^{2} + a x} b^{2}} - \frac {4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x}{\sqrt {b x^{2} + a x} b} + \frac {15 \, {\left (4 \, B a b^{3} + A b^{4}\right )} a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {5}{2}}} + \frac {2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(3/2),x, algorithm="maxima")

[Out]

1/3*B*b^3*x^4/sqrt(b*x^2 + a*x) - 7/12*B*a*b^2*x^3/sqrt(b*x^2 + a*x) + 35/24*B*a^2*b*x^2/sqrt(b*x^2 + a*x) + 5
1/8*B*a^3*x/sqrt(b*x^2 + a*x) + 4*A*a^2*b*x/sqrt(b*x^2 + a*x) - 35/16*B*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x
)*sqrt(b))/sqrt(b) - 2*A*a^3/sqrt(b*x^2 + a*x) + 1/2*(4*B*a*b^3 + A*b^4)*x^3/(sqrt(b*x^2 + a*x)*b) - 5/4*(4*B*
a*b^3 + A*b^4)*a*x^2/(sqrt(b*x^2 + a*x)*b^2) + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*x^2/(sqrt(b*x^2 + a*x)*b) - 15/4*(4
*B*a*b^3 + A*b^4)*a^2*x/(sqrt(b*x^2 + a*x)*b^3) + 6*(3*B*a^2*b^2 + 2*A*a*b^3)*a*x/(sqrt(b*x^2 + a*x)*b^2) - 4*
(2*B*a^3*b + 3*A*a^2*b^2)*x/(sqrt(b*x^2 + a*x)*b) + 15/8*(4*B*a*b^3 + A*b^4)*a^2*log(2*b*x + a + 2*sqrt(b*x^2
+ a*x)*sqrt(b))/b^(7/2) - 3*(3*B*a^2*b^2 + 2*A*a*b^3)*a*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) +
2*(2*B*a^3*b + 3*A*a^2*b^2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2)

Giac [A] (verification not implemented)

none

Time = 75.89 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\frac {{\left (\frac {{\left ({\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B}{b} + \frac {B a b + 6 \, A b^{2}}{b^{2}}\right )} + \frac {5 \, {\left (B a^{2} b + 6 \, A a b^{2}\right )}}{b^{2}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (B a^{3} b + 6 \, A a^{2} b^{2}\right )}}{b^{2}}\right )} \sqrt {b x + a}}{\sqrt {{\left (b x + a\right )} b - a b}} - \frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}}\right )} b^{2}}{24 \, {\left | b \right |}} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(3/2),x, algorithm="giac")

[Out]

1/24*(((2*(b*x + a)*(4*(b*x + a)*B/b + (B*a*b + 6*A*b^2)/b^2) + 5*(B*a^2*b + 6*A*a*b^2)/b^2)*(b*x + a) - 15*(B
*a^3*b + 6*A*a^2*b^2)/b^2)*sqrt(b*x + a)/sqrt((b*x + a)*b - a*b) - 15*(B*a^3 + 6*A*a^2*b)*log(abs(-sqrt(b*x +
a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(3/2))*b^2/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{x^{3/2}} \,d x \]

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^(3/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(5/2))/x^(3/2), x)