\(\int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx\) [659]

Optimal result

Integrand size = 22, antiderivative size = 299 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\frac {\left (64 a^2 b c d^2+(b c-5 a d) (b c-a d) (3 b c+a d)\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^2}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-2 a^{5/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{5/2}} \]

[Out]

1/24*(5*a*d+3*b*c)*(b*x+a)^(3/2)*(d*x+c)^(3/2)/d+1/4*(b*x+a)^(5/2)*(d*x+c)^(3/2)-2*a^(5/2)*c^(3/2)*arctanh(c^(
1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))+1/64*(-5*a^4*d^4+60*a^3*b*c*d^3+90*a^2*b^2*c^2*d^2-20*a*b^3*c^3*d+3*
b^4*c^4)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2)/d^(5/2)-1/32*(-5*a*d+b*c)*(a*d+3*b*c)*(d
*x+c)^(3/2)*(b*x+a)^(1/2)/d^2+1/64*(64*a^2*b*c*d^2+(-5*a*d+b*c)*(-a*d+b*c)*(a*d+3*b*c))*(b*x+a)^(1/2)*(d*x+c)^
(1/2)/b/d^2

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {103, 159, 163, 65, 223, 212, 95, 214} \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=-2 a^{5/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {1}{64} \sqrt {a+b x} \sqrt {c+d x} \left (\frac {5 a^3 d}{b}+73 a^2 c-\frac {17 a b c^2}{d}+\frac {3 b^2 c^3}{d^2}\right )+\frac {\left (-5 a^4 d^4+60 a^3 b c d^3+90 a^2 b^2 c^2 d^2-20 a b^3 c^3 d+3 b^4 c^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{5/2}}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (b c-5 a d) (a d+3 b c)}{32 d^2}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}+\frac {(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{24 d} \]

[In]

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

[Out]

((73*a^2*c + (3*b^2*c^3)/d^2 - (17*a*b*c^2)/d + (5*a^3*d)/b)*Sqrt[a + b*x]*Sqrt[c + d*x])/64 - ((b*c - 5*a*d)*
(3*b*c + a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(32*d^2) + ((3*b*c + 5*a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*
d) + ((a + b*x)^(5/2)*(c + d*x)^(3/2))/4 - 2*a^(5/2)*c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c +
d*x])] + ((3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(5/2))

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b
*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + p + 1))), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

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 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]

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 {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-\frac {1}{4} \int \frac {(a+b x)^{3/2} \sqrt {c+d x} \left (-4 a c+\frac {1}{2} (-3 b c-5 a d) x\right )}{x} \, dx \\ & = \frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-\frac {\int \frac {\sqrt {a+b x} \sqrt {c+d x} \left (-12 a^2 c d+\frac {3}{4} (b c-5 a d) (3 b c+a d) x\right )}{x} \, dx}{12 d} \\ & = -\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-\frac {\int \frac {\sqrt {c+d x} \left (-24 a^3 c d^2-\frac {3}{8} \left (64 a^2 b c d^2+(b c-5 a d) (b c-a d) (3 b c+a d)\right ) x\right )}{x \sqrt {a+b x}} \, dx}{24 d^2} \\ & = \frac {1}{64} \left (73 a^2 c+\frac {3 b^2 c^3}{d^2}-\frac {17 a b c^2}{d}+\frac {5 a^3 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-\frac {\int \frac {-24 a^3 b c^2 d^2-\frac {3}{16} \left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{24 b d^2} \\ & = \frac {1}{64} \left (73 a^2 c+\frac {3 b^2 c^3}{d^2}-\frac {17 a b c^2}{d}+\frac {5 a^3 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}+\left (a^3 c^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b d^2} \\ & = \frac {1}{64} \left (73 a^2 c+\frac {3 b^2 c^3}{d^2}-\frac {17 a b c^2}{d}+\frac {5 a^3 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}+\left (2 a^3 c^2\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{64 b^2 d^2} \\ & = \frac {1}{64} \left (73 a^2 c+\frac {3 b^2 c^3}{d^2}-\frac {17 a b c^2}{d}+\frac {5 a^3 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-2 a^{5/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{64 b^2 d^2} \\ & = \frac {1}{64} \left (73 a^2 c+\frac {3 b^2 c^3}{d^2}-\frac {17 a b c^2}{d}+\frac {5 a^3 d}{b}\right ) \sqrt {a+b x} \sqrt {c+d x}-\frac {(b c-5 a d) (3 b c+a d) \sqrt {a+b x} (c+d x)^{3/2}}{32 d^2}+\frac {(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 d}+\frac {1}{4} (a+b x)^{5/2} (c+d x)^{3/2}-2 a^{5/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^3 d^3+a^2 b d^2 (337 c+118 d x)+a b^2 d \left (57 c^2+244 c d x+136 d^2 x^2\right )+b^3 \left (-9 c^3+6 c^2 d x+72 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b d^2}-2 a^{5/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (3 b^4 c^4-20 a b^3 c^3 d+90 a^2 b^2 c^2 d^2+60 a^3 b c d^3-5 a^4 d^4\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{5/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^3*d^3 + a^2*b*d^2*(337*c + 118*d*x) + a*b^2*d*(57*c^2 + 244*c*d*x + 136*d^2
*x^2) + b^3*(-9*c^3 + 6*c^2*d*x + 72*c*d^2*x^2 + 48*d^3*x^3)))/(192*b*d^2) - 2*a^(5/2)*c^(3/2)*ArcTanh[(Sqrt[c
]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3
- 5*a^4*d^4)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(708\) vs. \(2(249)=498\).

Time = 1.59 (sec) , antiderivative size = 709, normalized size of antiderivative = 2.37

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-96 b^{3} d^{3} x^{3} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-272 a \,b^{2} d^{3} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}-144 b^{3} c \,d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{4} d^{4}-180 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{3} b c \,d^{3}-270 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} b^{2} c^{2} d^{2}+60 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a \,b^{3} c^{3} d -9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{4} c^{4}+384 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{3} b \,c^{2} d^{2}-236 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b \,d^{3} x -488 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c \,d^{2} x -12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{2} d x -30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{3} d^{3}-674 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} b c \,d^{2}-114 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} c^{2} d +18 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c^{3}\right )}{384 b \,d^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}}\)	\(709\)

[In]

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

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*b^3*d^3*x^3*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-272*a*b^2*
d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-144*b^3*c*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*
(a*c)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*a^4*d^4
-180*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*a^3*b*c*d^3-270*l
n(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*a^2*b^2*c^2*d^2+60*ln(1
/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*a*b^3*c^3*d-9*ln(1/2*(2*b*
d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*b^4*c^4+384*(b*d)^(1/2)*ln((a*d*x+
b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*a^3*b*c^2*d^2-236*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a
*c)^(1/2)*a^2*b*d^3*x-488*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b^2*c*d^2*x-12*((b*x+a)*(d*x+c))^(
1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^3*c^2*d*x-30*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^3*d^3-674*((b*x+
a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*b*c*d^2-114*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b^
2*c^2*d+18*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^3*c^3)/b/d^2/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/
(a*c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 8.55 (sec) , antiderivative size = 1481, normalized size of antiderivative = 4.95 \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/768*(384*sqrt(a*c)*a^2*b^2*c*d^3*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a
*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 3*(3*b^4*c^4 - 20*a*b^3*c^3*d +
90*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
- 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(48*b^4*d^4*x^3
- 9*b^4*c^3*d + 57*a*b^3*c^2*d^2 + 337*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(9*b^4*c*d^3 + 17*a*b^3*d^4)*x^2 + 2*
(3*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^3), 1/384*(192*sqrt(
a*c)*a^2*b^2*c*d^3*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*
sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 3*(3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d
^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x
+ c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(48*b^4*d^4*x^3 - 9*b^4*c^3*d + 57*a*b^3*c^2*d^2 +
337*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(9*b^4*c*d^3 + 17*a*b^3*d^4)*x^2 + 2*(3*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 5
9*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^3), 1/768*(768*sqrt(-a*c)*a^2*b^2*c*d^3*arctan(1/2*(2*a*
c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 3
*(3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*sqrt(b*d)*log(8*b^2*d^2*x^2 +
b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a
*b*d^2)*x) + 4*(48*b^4*d^4*x^3 - 9*b^4*c^3*d + 57*a*b^3*c^2*d^2 + 337*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(9*b^4*
c*d^3 + 17*a*b^3*d^4)*x^2 + 2*(3*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c
))/(b^2*d^3), 1/384*(384*sqrt(-a*c)*a^2*b^2*c*d^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*
sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 3*(3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^
2*d^2 + 60*a^3*b*c*d^3 - 5*a^4*d^4)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(
d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(48*b^4*d^4*x^3 - 9*b^4*c^3*d + 57*a*b^3*c^2*d^2
+ 337*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(9*b^4*c*d^3 + 17*a*b^3*d^4)*x^2 + 2*(3*b^4*c^2*d^2 + 122*a*b^3*c*d^3
+ 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^3)]

Sympy [F]

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

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(3/2)/x,x)

[Out]

Integral((a + b*x)**(5/2)*(c + d*x)**(3/2)/x, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{5/2} (c+d x)^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

int(((a + b*x)^(5/2)*(c + d*x)^(3/2))/x,x)

[Out]

int(((a + b*x)^(5/2)*(c + d*x)^(3/2))/x, x)