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

Optimal. Leaf size=171 \[ \frac {\left (-15 a^2 d^2+6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} c^{7/2}}-\frac {d \sqrt {a+b x} (b c-15 a d)}{4 a c^3 \sqrt {c+d x}}-\frac {\sqrt {a+b x} (b c-5 a d)}{4 a c^2 x \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}} \]

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Rubi [A]  time = 0.13, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {99, 151, 152, 12, 93, 208} \begin {gather*} \frac {\left (-15 a^2 d^2+6 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} c^{7/2}}-\frac {d \sqrt {a+b x} (b c-15 a d)}{4 a c^3 \sqrt {c+d x}}-\frac {\sqrt {a+b x} (b c-5 a d)}{4 a c^2 x \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(b*c - 15*a*d)*Sqrt[a + b*x])/(4*a*c^3*Sqrt[c + d*x]) - Sqrt[a + b*x]/(2*c*x^2*Sqrt[c + d*x]) - ((b*c - 5*
a*d)*Sqrt[a + b*x])/(4*a*c^2*x*Sqrt[c + d*x]) + ((b^2*c^2 + 6*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)*c^(7/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 99

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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x}}{x^3 (c+d x)^{3/2}} \, dx &=-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}}+\frac {\int \frac {\frac {1}{2} (b c-5 a d)-2 b d x}{x^2 \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{2 c}\\ &=-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}}-\frac {(b c-5 a d) \sqrt {a+b x}}{4 a c^2 x \sqrt {c+d x}}-\frac {\int \frac {\frac {1}{4} \left (b^2 c^2+6 a b c d-15 a^2 d^2\right )+\frac {1}{2} b d (b c-5 a d) x}{x \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{2 a c^2}\\ &=-\frac {d (b c-15 a d) \sqrt {a+b x}}{4 a c^3 \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}}-\frac {(b c-5 a d) \sqrt {a+b x}}{4 a c^2 x \sqrt {c+d x}}+\frac {\int -\frac {(b c-a d) \left (b^2 c^2+6 a b c d-15 a^2 d^2\right )}{8 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{a c^3 (b c-a d)}\\ &=-\frac {d (b c-15 a d) \sqrt {a+b x}}{4 a c^3 \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}}-\frac {(b c-5 a d) \sqrt {a+b x}}{4 a c^2 x \sqrt {c+d x}}-\frac {\left (b^2 c^2+6 a b c d-15 a^2 d^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 a c^3}\\ &=-\frac {d (b c-15 a d) \sqrt {a+b x}}{4 a c^3 \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}}-\frac {(b c-5 a d) \sqrt {a+b x}}{4 a c^2 x \sqrt {c+d x}}-\frac {\left (b^2 c^2+6 a b c d-15 a^2 d^2\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 c^3}\\ &=-\frac {d (b c-15 a d) \sqrt {a+b x}}{4 a c^3 \sqrt {c+d x}}-\frac {\sqrt {a+b x}}{2 c x^2 \sqrt {c+d x}}-\frac {(b c-5 a d) \sqrt {a+b x}}{4 a c^2 x \sqrt {c+d x}}+\frac {\left (b^2 c^2+6 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} c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 130, normalized size = 0.76 \begin {gather*} \frac {\left (-15 a^2 d^2+6 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} c^{7/2}}+\frac {\sqrt {a+b x} \left (a \left (-2 c^2+5 c d x+15 d^2 x^2\right )-b c x (c+d x)\right )}{4 a c^3 x^2 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-(b*c*x*(c + d*x)) + a*(-2*c^2 + 5*c*d*x + 15*d^2*x^2)))/(4*a*c^3*x^2*Sqrt[c + d*x]) + ((b^2*c
^2 + 6*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)*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.28, size = 224, normalized size = 1.31 \begin {gather*} \frac {\left (-15 a^2 d^2+6 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} c^{7/2}}+\frac {\sqrt {a+b x} \left (15 a^3 d^2-\frac {25 a^2 c d^2 (a+b x)}{c+d x}-6 a^2 b c d-\frac {b^2 c^3 (a+b x)}{c+d x}-a b^2 c^2+\frac {8 a c^2 d^2 (a+b x)^2}{(c+d x)^2}+\frac {10 a b c^2 d (a+b x)}{c+d x}\right )}{4 a c^3 \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(-(a*b^2*c^2) - 6*a^2*b*c*d + 15*a^3*d^2 + (8*a*c^2*d^2*(a + b*x)^2)/(c + d*x)^2 - (b^2*c^3*(a
+ b*x))/(c + d*x) + (10*a*b*c^2*d*(a + b*x))/(c + d*x) - (25*a^2*c*d^2*(a + b*x))/(c + d*x)))/(4*a*c^3*Sqrt[c
+ d*x]*(a - (c*(a + b*x))/(c + d*x))^2) + ((b^2*c^2 + 6*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)*c^(7/2))

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fricas [A]  time = 3.28, size = 474, normalized size = 2.77 \begin {gather*} \left [-\frac {{\left ({\left (b^{2} c^{2} d + 6 \, a b c d^{2} - 15 \, a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} + 6 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + {\left (a b c^{3} - 5 \, a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a^{2} c^{4} d x^{3} + a^{2} c^{5} x^{2}\right )}}, -\frac {{\left ({\left (b^{2} c^{2} d + 6 \, a b c d^{2} - 15 \, a^{2} d^{3}\right )} x^{3} + {\left (b^{2} c^{3} + 6 \, a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (2 \, a^{2} c^{3} + {\left (a b c^{2} d - 15 \, a^{2} c d^{2}\right )} x^{2} + {\left (a b c^{3} - 5 \, a^{2} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a^{2} c^{4} d x^{3} + a^{2} c^{5} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(((b^2*c^2*d + 6*a*b*c*d^2 - 15*a^2*d^3)*x^3 + (b^2*c^3 + 6*a*b*c^2*d - 15*a^2*c*d^2)*x^2)*sqrt(a*c)*lo
g((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) + 4*(2*a^2*c^3 + (a*b*c^2*d - 15*a^2*c*d^2)*x^2 + (a*b*c^3 - 5*a^2*c^2*
d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*d*x^3 + a^2*c^5*x^2), -1/8*(((b^2*c^2*d + 6*a*b*c*d^2 - 15*a^2*d^3
)*x^3 + (b^2*c^3 + 6*a*b*c^2*d - 15*a^2*c*d^2)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*s
qrt(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)) + 2*(2*a^2*c^3 + (a*b*c^2*d - 15*a
^2*c*d^2)*x^2 + (a*b*c^3 - 5*a^2*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^4*d*x^3 + a^2*c^5*x^2)]

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giac [B]  time = 15.39, size = 1092, normalized size = 6.39 \begin {gather*} \frac {2 \, \sqrt {b x + a} b^{2} d^{2}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} c^{3} {\left | b \right |}} + \frac {{\left (\sqrt {b d} b^{4} c^{2} + 6 \, \sqrt {b d} a b^{3} c d - 15 \, \sqrt {b d} a^{2} b^{2} d^{2}\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 )}{4 \, \sqrt {-a b c d} a b c^{3} {\left | b \right |}} - \frac {\sqrt {b d} b^{10} c^{5} - 11 \, \sqrt {b d} a b^{9} c^{4} d + 34 \, \sqrt {b d} a^{2} b^{8} c^{3} d^{2} - 46 \, \sqrt {b d} a^{3} b^{7} c^{2} d^{3} + 29 \, \sqrt {b d} a^{4} b^{6} c d^{4} - 7 \, \sqrt {b d} a^{5} b^{5} d^{5} - 3 \, \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^{8} c^{4} + 28 \, \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^{7} c^{3} d - 26 \, \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^{2} b^{6} c^{2} d^{2} - 20 \, \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^{3} b^{5} c d^{3} + 21 \, \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^{4} b^{4} d^{4} + 3 \, \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 )}^{4} b^{6} c^{3} - 19 \, \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 )}^{4} a b^{5} c^{2} d - 11 \, \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 )}^{4} a^{2} b^{4} c d^{2} - 21 \, \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 )}^{4} a^{3} b^{3} d^{3} - \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 )}^{6} b^{4} c^{2} + 2 \, \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 )}^{6} a b^{3} c d + 7 \, \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 )}^{6} a^{2} b^{2} d^{2}}{2 \, {\left (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}\right )}^{2} a c^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + a)*b^2*d^2/(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*c^3*abs(b)) + 1/4*(sqrt(b*d)*b^4*c^2 + 6*sqrt(b*d
)*a*b^3*c*d - 15*sqrt(b*d)*a^2*b^2*d^2)*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*c^3*abs(b)) - 1/2*(sqrt(b*d)*b^10*c^5 - 11*s
qrt(b*d)*a*b^9*c^4*d + 34*sqrt(b*d)*a^2*b^8*c^3*d^2 - 46*sqrt(b*d)*a^3*b^7*c^2*d^3 + 29*sqrt(b*d)*a^4*b^6*c*d^
4 - 7*sqrt(b*d)*a^5*b^5*d^5 - 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^
8*c^4 + 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*b^7*c^3*d - 26*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^6*c^2*d^2 - 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^3*b^5*c*d^3 + 21*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^4*d^4 + 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^6*c^3 - 19*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^5*c^2*d - 11*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^4*c*d^2
- 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^3*b^3*d^3 - sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c^2 + 2*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c*d + 7*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^6*a^2*b^2*d^2)/((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*c^3*abs(b))

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maple [B]  time = 0.03, size = 467, normalized size = 2.73 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (15 a^{2} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 a b c \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b^{2} c^{2} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+15 a^{2} c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-6 a b \,c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-b^{2} c^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,d^{2} x^{2}+2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c d \,x^{2}-10 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c d x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2} x +4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,c^{2}\right )}{8 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {d x +c}\, a \,c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b*x+a)^(1/2)/a/c^3*(15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^2*d^3-6*ln(
(a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a*b*c*d^2-ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)
*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*b^2*c^2*d+15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*
x^2*a^2*c*d^2-6*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*b*c^2*d-ln((a*d*x+b*c*x+
2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*b^2*c^3-30*x^2*a*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2
*x^2*b*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*a*c*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*x*b*c^2*(a*c)^
(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/(a*c)^(1/2)/x^2/((b*x+a)*(d*x+c))^(
1/2)/(d*x+c)^(1/2)

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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((b*x+a)^(1/2)/x^3/(d*x+c)^(3/2),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 \begin {gather*} \int \frac {\sqrt {a+b\,x}}{x^3\,{\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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