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

Optimal. Leaf size=211 \[ \frac {\sqrt {c} \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}+\frac {d^{3/2} (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{4 a x}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{4 a} \]

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {97, 149, 154, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {\sqrt {c} \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}+\frac {d^{3/2} (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}-\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+b c)}{4 a x}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (11 a d+b c)}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*c + 11*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*a) - ((b*c + 5*a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(4*a*x) -
(Sqrt[a + b*x]*(c + d*x)^(5/2))/(2*x^2) + (Sqrt[c]*(b^2*c^2 - 10*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)) + (d^(3/2)*(5*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt
[b]*Sqrt[c + d*x])])/Sqrt[b]

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 93

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 97

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

Rule 149

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

Rule 154

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 157

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 206

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

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]

Rule 217

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*} \int \frac {\sqrt {a+b x} (c+d x)^{5/2}}{x^3} \, dx &=-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {1}{2} \int \frac {(c+d x)^{3/2} \left (\frac {1}{2} (b c+5 a d)+3 b d x\right )}{x^2 \sqrt {a+b x}} \, dx\\ &=-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{4} \left (-b^2 c^2+10 a b c d+15 a^2 d^2\right )+\frac {1}{2} b d (b c+11 a d) x\right )}{x \sqrt {a+b x}} \, dx}{2 a}\\ &=\frac {d (b c+11 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {-\frac {1}{4} b c \left (b^2 c^2-10 a b c d-15 a^2 d^2\right )+a b d^2 (5 b c+a d) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 a b}\\ &=\frac {d (b c+11 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {1}{2} \left (d^2 (5 b c+a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx-\frac {\left (c \left (b^2 c^2-10 a b c d-15 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 a}\\ &=\frac {d (b c+11 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {\left (d^2 (5 b c+a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{b}-\frac {\left (c \left (b^2 c^2-10 a b c d-15 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 a}\\ &=\frac {d (b c+11 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {\sqrt {c} \left (b^2 c^2-10 a b c d-15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}+\frac {\left (d^2 (5 b c+a d)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{b}\\ &=\frac {d (b c+11 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 a}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{4 a x}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{2 x^2}+\frac {\sqrt {c} \left (b^2 c^2-10 a b c d-15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}+\frac {d^{3/2} (5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.40, size = 207, normalized size = 0.98 \begin {gather*} \frac {\sqrt {c} \left (-15 a^2 d^2-10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a \left (2 c^2+9 c d x-4 d^2 x^2\right )+b c^2 x\right )}{4 a x^2}+\frac {d^{3/2} \sqrt {c+d x} (a d+5 b c) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[a + b*x]*Sqrt[c + d*x]*(b*c^2*x + a*(2*c^2 + 9*c*d*x - 4*d^2*x^2)))/(a*x^2) + (d^(3/2)*(5*b*c + a*d
)*Sqrt[c + d*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x))/(b*c - a
*d)]) + (Sqrt[c]*(b^2*c^2 - 10*a*b*c*d - 15*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])
/(4*a^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.72, size = 387, normalized size = 1.83 \begin {gather*} \frac {\left (-15 a^2 \sqrt {c} d^2-10 a b c^{3/2} d+b^2 c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2}}+\frac {\sqrt {a+b x} \left (4 a^4 d^3-\frac {17 a^3 c d^3 (a+b x)}{c+d x}+5 a^3 b c d^2-10 a^2 b^2 c^2 d+\frac {11 a^2 c^2 d^3 (a+b x)^2}{(c+d x)^2}+\frac {11 a^2 b c^2 d^2 (a+b x)}{c+d x}+\frac {b^3 c^4 (a+b x)}{c+d x}+a b^3 c^3-\frac {b^2 c^4 d (a+b x)^2}{(c+d x)^2}+\frac {5 a b^2 c^3 d (a+b x)}{c+d x}-\frac {10 a b c^3 d^2 (a+b x)^2}{(c+d x)^2}\right )}{4 a \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right )^2 \left (\frac {d (a+b x)}{c+d x}-b\right )}+\frac {\left (a d^{5/2}+5 b c d^{3/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(a*b^3*c^3 - 10*a^2*b^2*c^2*d + 5*a^3*b*c*d^2 + 4*a^4*d^3 - (b^2*c^4*d*(a + b*x)^2)/(c + d*x)^2
- (10*a*b*c^3*d^2*(a + b*x)^2)/(c + d*x)^2 + (11*a^2*c^2*d^3*(a + b*x)^2)/(c + d*x)^2 + (b^3*c^4*(a + b*x))/(
c + d*x) + (5*a*b^2*c^3*d*(a + b*x))/(c + d*x) + (11*a^2*b*c^2*d^2*(a + b*x))/(c + d*x) - (17*a^3*c*d^3*(a + b
*x))/(c + d*x)))/(4*a*Sqrt[c + d*x]*(a - (c*(a + b*x))/(c + d*x))^2*(-b + (d*(a + b*x))/(c + d*x))) + ((b^2*c^
(5/2) - 10*a*b*c^(3/2)*d - 15*a^2*Sqrt[c]*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^
(3/2)) + ((5*b*c*d^(3/2) + a*d^(5/2))*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]

________________________________________________________________________________________

fricas [A]  time = 8.06, size = 1094, normalized size = 5.18

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/16*(4*(5*a*b*c*d + a^2*d^2)*x^2*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x
+ b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - (b^2*c^2 - 10*a*b*c*d - 15
*a^2*d^2)*x^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)
*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(4*a*d^2*x^2 - 2*a*c^2 - (b*c^2 + 9
*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2), -1/16*(8*(5*a*b*c*d + a^2*d^2)*x^2*sqrt(-d/b)*arctan(1/2*(2*b
*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + (b^2*c^2 -
10*a*b*c*d - 15*a^2*d^2)*x^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a
*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(4*a*d^2*x^2 - 2*a*
c^2 - (b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2), -1/8*((b^2*c^2 - 10*a*b*c*d - 15*a^2*d^2)*x^2
*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*
c^2 + a*c*d)*x)) - 2*(5*a*b*c*d + a^2*d^2)*x^2*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4
*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - 2*(4*a*d^2*x^2
- 2*a*c^2 - (b*c^2 + 9*a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2), -1/8*((b^2*c^2 - 10*a*b*c*d - 15*a^2*d
^2)*x^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^
2 + (b*c^2 + a*c*d)*x)) + 4*(5*a*b*c*d + a^2*d^2)*x^2*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a
)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - 2*(4*a*d^2*x^2 - 2*a*c^2 - (b*c^2 + 9*a*
c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*x^2)]

________________________________________________________________________________________

giac [B]  time = 16.31, size = 1200, normalized size = 5.69

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/4*(4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d^2*abs(b)/b - 2*(5*sqrt(b*d)*b*c*d*abs(b) + sqrt(b*d
)*a*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b + (sqrt(b*d)*b^3*c^3*
abs(b) - 10*sqrt(b*d)*a*b^2*c^2*d*abs(b) - 15*sqrt(b*d)*a^2*b*c*d^2*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a*b) - 2*(sq
rt(b*d)*b^9*c^6*abs(b) + 5*sqrt(b*d)*a*b^8*c^5*d*abs(b) - 30*sqrt(b*d)*a^2*b^7*c^4*d^2*abs(b) + 50*sqrt(b*d)*a
^3*b^6*c^3*d^3*abs(b) - 35*sqrt(b*d)*a^4*b^5*c^2*d^4*abs(b) + 9*sqrt(b*d)*a^5*b^4*c*d^5*abs(b) - 3*sqrt(b*d)*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^5*abs(b) - 20*sqrt(b*d)*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^4*d*abs(b) + 22*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*c^3*d^2*abs(b) + 28*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt
(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c^2*d^3*abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^2*a^4*b^3*c*d^4*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^4*b^5*c^4*abs(b) + 29*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^4*a*b^4*c^3*d*abs(b) + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*
b^3*c^2*d^2*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^2*c*
d^3*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^3*c^3*abs(b) - 14*s
qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^2*c^2*d*abs(b) - 9*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*c*d^2*abs(b))/((b^4*c^2 - 2*a*b^3*c*d +
a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*
x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4)^2*a))/b

________________________________________________________________________________________

maple [B]  time = 0.02, size = 511, normalized size = 2.42 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-15 \sqrt {b d}\, a^{2} c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+4 \sqrt {a c}\, a^{2} d^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-10 \sqrt {b d}\, a b \,c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+20 \sqrt {a c}\, a b c \,d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+\sqrt {b d}\, b^{2} c^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,d^{2} x^{2}-18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a c d x -2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b \,c^{2} x -4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,c^{2}\right )}{8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3,x)

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a*(4*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b
*d)^(1/2))*x^2*a^2*d^3*(a*c)^(1/2)+20*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(
b*d)^(1/2))*x^2*a*b*c*d^2*(a*c)^(1/2)-15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/
x)*x^2*a^2*c*d^2*(b*d)^(1/2)-10*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*a*
b*c^2*d*(b*d)^(1/2)+ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*b^2*c^3*(b*d)^
(1/2)+8*x^2*a*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-18*x*a*c*d*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-2*x*b*c^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-4*a*c^2*(b*d*
x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x^2/(b*d)^(1/2)/(a*c)^(1/2
)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*(b*x+a)^(1/2)/x^3,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 details)Is a*d-b*c zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^3,x)

[Out]

int(((a + b*x)^(1/2)*(c + d*x)^(5/2))/x^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)*(b*x+a)**(1/2)/x**3,x)

[Out]

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

________________________________________________________________________________________