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

Optimal. Leaf size=198 \[ \frac {\sqrt {d} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2}}-\frac {c^{3/2} (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}+\frac {3}{2} d \sqrt {a+b x} (c+d x)^{3/2}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{4 b} \]

[Out]

-c^(3/2)*(5*a*d+b*c)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(1/2)+1/4*(-a^2*d^2+10*a*b*c*d+15*
b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*d^(1/2)/b^(3/2)+3/2*d*(d*x+c)^(3/2)*(b*x+a)^(1/2
)-(d*x+c)^(5/2)*(b*x+a)^(1/2)/x+1/4*d*(a*d+11*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b

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Rubi [A]  time = 0.20, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {97, 154, 157, 63, 217, 206, 93, 208} \[ \frac {\sqrt {d} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{3/2}}-\frac {c^{3/2} (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x} (c+d x)^{5/2}}{x}+\frac {3}{2} d \sqrt {a+b x} (c+d x)^{3/2}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+11 b c)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 1.31, size = 208, normalized size = 1.05 \[ \frac {\frac {\sqrt {d} \sqrt {c+d x} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}}-\frac {4 b c^{3/2} (5 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a d^2 x+b \left (-4 c^2+9 c d x+2 d^2 x^2\right )\right )}{x}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 5.87, size = 1079, normalized size = 5.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*x*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) - 4*(b^2*c^2 + 5*
a*b*c*d)*x*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)*sq
rt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*b*d^2*x^2 - 4*b*c^2 + (9*b*c*d + a*
d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*x), -1/8*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*x*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*(b^2
*c^2 + 5*a*b*c*d)*x*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) - 2*(2*b*d^2*x^2 - 4*b*c^2 + (9*b
*c*d + a*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*x), 1/16*(8*(b^2*c^2 + 5*a*b*c*d)*x*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)) - (15*b^2
*c^2 + 10*a*b*c*d - a^2*d^2)*x*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) + 4*(2*b*d^2*x^2 - 4*b*c^2 + (9*
b*c*d + a*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*x), 1/8*(4*(b^2*c^2 + 5*a*b*c*d)*x*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)) - (15*b^2
*c^2 + 10*a*b*c*d - a^2*d^2)*x*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*(2*b*d^2*x^2 - 4*b*c^2 + (9*b*c*d + a*d^2)*x)*sqrt(b*x + a)*s
qrt(d*x + c))/(b*x)]

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giac [B]  time = 4.06, size = 597, normalized size = 3.02 \[ \frac {2 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{2}} + \frac {9 \, b^{3} c d^{3} {\left | b \right |} - a b^{2} d^{4} {\left | b \right |}}{b^{4} d^{2}}\right )} - \frac {8 \, {\left (\sqrt {b d} b^{2} c^{3} {\left | b \right |} + 5 \, \sqrt {b d} a b c^{2} d {\left | b \right |}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b} - \frac {16 \, {\left (\sqrt {b d} b^{4} c^{4} {\left | b \right |} - 2 \, \sqrt {b d} a b^{3} c^{3} d {\left | b \right |} + \sqrt {b d} a^{2} b^{2} c^{2} d^{2} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c^{3} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b c^{2} d {\left | b \right |}\right )}}{b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}} - \frac {{\left (15 \, \sqrt {b d} b^{2} c^{2} {\left | b \right |} + 10 \, \sqrt {b d} a b c d {\left | b \right |} - \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b^{2}}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*d^2*abs(b)/b^2 + (9*b^3*c*d^3*abs(b) - a
*b^2*d^4*abs(b))/(b^4*d^2)) - 8*(sqrt(b*d)*b^2*c^3*abs(b) + 5*sqrt(b*d)*a*b*c^2*d*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
)*b) - 16*(sqrt(b*d)*b^4*c^4*abs(b) - 2*sqrt(b*d)*a*b^3*c^3*d*abs(b) + sqrt(b*d)*a^2*b^2*c^2*d^2*abs(b) - sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c^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))^2*a*b*c^2*d*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) - sqr
t(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)
- (15*sqrt(b*d)*b^2*c^2*abs(b) + 10*sqrt(b*d)*a*b*c*d*abs(b) - sqrt(b*d)*a^2*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^2)/b

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maple [B]  time = 0.02, size = 503, normalized size = 2.54 \[ -\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (\sqrt {a c}\, a^{2} d^{3} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+20 \sqrt {b d}\, a b \,c^{2} d x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-10 \sqrt {a c}\, a b c \,d^{2} x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {b d}\, b^{2} c^{3} x \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-15 \sqrt {a c}\, b^{2} c^{2} d x \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b \,d^{2} x^{2}-2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,d^{2} x -18 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b c d x +8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b \,c^{2}\right )}{8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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