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

Optimal. Leaf size=131 \[ -\frac {(b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{4 a^2 c x}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \begin {gather*} -\frac {(b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (a d+3 b c)}{4 a^2 c x}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 a c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 94

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[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

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Mathematica [A]  time = 0.09, size = 112, normalized size = 0.85 \begin {gather*} \frac {\left (a^2 d^2+2 a b c d-3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (2 a c+a d x-3 b c x)}{4 a^2 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.48, size = 241, normalized size = 1.84 \begin {gather*} \frac {\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (-a c d^2 \sqrt {c+d x}-a d^2 (c+d x)^{3/2}-3 b c^2 d \sqrt {c+d x}+3 b c d (c+d x)^{3/2}\right )}{4 a^2 c d^2 x^2}-\frac {\sqrt {\frac {b}{d}} \left (-a^2 d^{5/2}-2 a b c d^{3/2}+3 b^2 c^2 \sqrt {d}\right ) \tanh ^{-1}\left (\frac {d \sqrt {\frac {b}{d}} \sqrt {c+d x} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}-b (c+d x)+b c}{\sqrt {a} \sqrt {b} \sqrt {c} \sqrt {d}}\right )}{4 a^{5/2} \sqrt {b} c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.64, size = 336, normalized size = 2.56 \begin {gather*} \left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {a c} x^{2} \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^{2} - {\left (3 \, a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{3} c^{2} x^{2}}, \frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \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^{2} - {\left (3 \, a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{3} c^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*sqrt(a*c)*x^2*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) + 4*(2*a^2*c^
2 - (3*a*b*c^2 - a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c^2*x^2), 1/8*((3*b^2*c^2 - 2*a*b*c*d - a^2*d^2
)*sqrt(-a*c)*x^2*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)) - 2*(2*a^2*c^2 - (3*a*b*c^2 - a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*c^2*
x^2)]

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giac [B]  time = 3.29, size = 1048, normalized size = 8.00 \begin {gather*} -\frac {{\left (\frac {{\left (3 \, \sqrt {b d} b^{5} c^{2} - 2 \, \sqrt {b d} a b^{4} c d - \sqrt {b d} a^{2} b^{3} 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 )}{\sqrt {-a b c d} a^{2} b c} - \frac {2 \, {\left (3 \, \sqrt {b d} b^{11} c^{5} - 13 \, \sqrt {b d} a b^{10} c^{4} d + 22 \, \sqrt {b d} a^{2} b^{9} c^{3} d^{2} - 18 \, \sqrt {b d} a^{3} b^{8} c^{2} d^{3} + 7 \, \sqrt {b d} a^{4} b^{7} c d^{4} - \sqrt {b d} a^{5} b^{6} d^{5} - 9 \, \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^{9} c^{4} + 8 \, \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^{8} c^{3} d + 14 \, \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^{7} c^{2} d^{2} - 16 \, \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^{6} c d^{3} + 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} a^{4} b^{5} d^{4} + 9 \, \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^{7} c^{3} + 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} a b^{6} c^{2} 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 )}^{4} a^{2} b^{5} c d^{2} - 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} a^{3} b^{4} d^{3} - 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^{5} 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^{4} c d + \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^{3} d^{2}\right )}}{{\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^{2} c}\right )} {\left | b \right |}}{4 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((3*sqrt(b*d)*b^5*c^2 - 2*sqrt(b*d)*a*b^4*c*d - sqrt(b*d)*a^2*b^3*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^2*b*c) - 2
*(3*sqrt(b*d)*b^11*c^5 - 13*sqrt(b*d)*a*b^10*c^4*d + 22*sqrt(b*d)*a^2*b^9*c^3*d^2 - 18*sqrt(b*d)*a^3*b^8*c^2*d
^3 + 7*sqrt(b*d)*a^4*b^7*c*d^4 - sqrt(b*d)*a^5*b^6*d^5 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^2*b^9*c^4 + 8*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^8*c^3*d + 14*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^7*c^2*d^
2 - 16*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^6*c*d^3 + 3*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^5*d^4 + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^7*c^3 + 3*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^6*c^2*d + 7*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^5*c*d^2 - 3*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^4
*d^3 - 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^5*c^2 + 2*sqrt(b*d)*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^4*c*d + 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^3*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqr
t(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^2*c))*abs(b)/
b^3

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maple [B]  time = 0.02, size = 257, normalized size = 1.96 \begin {gather*} \frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (a^{2} 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 )+2 a b c 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 )-3 b^{2} c^{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 )-2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a d x +6 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b c x -4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a c \right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^2/c*(a^2*d^2*x^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)+2*a*b*c*d*x^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)-3*b^2*c^2*x^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)-2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*d*x+6*((b*x+a)*(d*x
+c))^(1/2)*(a*c)^(1/2)*b*c*x-4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*c)/((b*x+a)*(d*x+c))^(1/2)/x^2/(a*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((d*x+c)^(1/2)/x^3/(b*x+a)^(1/2),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 [B]  time = 20.07, size = 901, normalized size = 6.88 \begin {gather*} \frac {\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+d\,x}-\sqrt {c}}\right )\,\left (a^{5/2}\,\sqrt {c}\,d^2-3\,\sqrt {a}\,b^2\,c^{5/2}+2\,a^{3/2}\,b\,c^{3/2}\,d\right )}{8\,a^3\,c^2}-\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {d^2}{4\,a\,c}-\frac {3\,d\,\left (a\,d+b\,c\right )}{16\,a^2\,c}\right )}{\sqrt {c+d\,x}-\sqrt {c}}-\frac {\ln \left (\frac {\left (\sqrt {c}\,\sqrt {a+b\,x}-\sqrt {a}\,\sqrt {c+d\,x}\right )\,\left (b\,\sqrt {c}-\frac {\sqrt {a}\,d\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {c+d\,x}-\sqrt {c}}\right )}{\sqrt {c+d\,x}-\sqrt {c}}\right )\,\left (a^{5/2}\,\sqrt {c}\,d^2-3\,\sqrt {a}\,b^2\,c^{5/2}+2\,a^{3/2}\,b\,c^{3/2}\,d\right )}{8\,a^3\,c^2}-\frac {\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2\,\left (-\frac {5\,a^2\,b^2\,d^2}{32}+\frac {a\,b^3\,c\,d}{8}+\frac {11\,b^4\,c^2}{32}\right )}{a^{5/2}\,c^{3/2}\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}-\frac {b^4}{32\,a^{3/2}\,\sqrt {c}\,d^2}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {a^3\,b\,d^3}{16}+\frac {3\,a^2\,b^2\,c\,d^2}{8}-\frac {9\,a\,b^3\,c^2\,d}{8}+\frac {b^4\,c^3}{16}\right )}{a^3\,c^2\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {\left (\frac {b^4\,c}{8}-\frac {a\,b^3\,d}{8}\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{a^2\,c\,d^2\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {5\,a^2\,b\,d^2}{16}-\frac {11\,a\,b^2\,c\,d}{16}+\frac {b^3\,c^2}{4}\right )}{a^3\,c\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (\frac {a^4\,d^4}{32}-\frac {a^3\,b\,c\,d^3}{2}+\frac {21\,a^2\,b^2\,c^2\,d^2}{32}+\frac {a\,b^3\,c^3\,d}{2}-\frac {7\,b^4\,c^4}{32}\right )}{a^{7/2}\,c^{5/2}\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )}{a\,c\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {\left (2\,c\,b^2+2\,a\,d\,b\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{\sqrt {a}\,\sqrt {c}\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {\left (2\,a\,d+2\,b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}{\sqrt {a}\,\sqrt {c}\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}}+\frac {d^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{32\,a^{3/2}\,\sqrt {c}\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(((a + b*x)^(1/2) - a^(1/2))/((c + d*x)^(1/2) - c^(1/2)))*(a^(5/2)*c^(1/2)*d^2 - 3*a^(1/2)*b^2*c^(5/2) + 2
*a^(3/2)*b*c^(3/2)*d))/(8*a^3*c^2) - (((a + b*x)^(1/2) - a^(1/2))*(d^2/(4*a*c) - (3*d*(a*d + b*c))/(16*a^2*c))
)/((c + d*x)^(1/2) - c^(1/2)) - (log(((c^(1/2)*(a + b*x)^(1/2) - a^(1/2)*(c + d*x)^(1/2))*(b*c^(1/2) - (a^(1/2
)*d*((a + b*x)^(1/2) - a^(1/2)))/((c + d*x)^(1/2) - c^(1/2))))/((c + d*x)^(1/2) - c^(1/2)))*(a^(5/2)*c^(1/2)*d
^2 - 3*a^(1/2)*b^2*c^(5/2) + 2*a^(3/2)*b*c^(3/2)*d))/(8*a^3*c^2) - ((((a + b*x)^(1/2) - a^(1/2))^2*((11*b^4*c^
2)/32 - (5*a^2*b^2*d^2)/32 + (a*b^3*c*d)/8))/(a^(5/2)*c^(3/2)*d^2*((c + d*x)^(1/2) - c^(1/2))^2) - b^4/(32*a^(
3/2)*c^(1/2)*d^2) + (((a + b*x)^(1/2) - a^(1/2))^3*((b^4*c^3)/16 + (a^3*b*d^3)/16 + (3*a^2*b^2*c*d^2)/8 - (9*a
*b^3*c^2*d)/8))/(a^3*c^2*d^2*((c + d*x)^(1/2) - c^(1/2))^3) - (((b^4*c)/8 - (a*b^3*d)/8)*((a + b*x)^(1/2) - a^
(1/2)))/(a^2*c*d^2*((c + d*x)^(1/2) - c^(1/2))) + (((a + b*x)^(1/2) - a^(1/2))^5*((b^3*c^2)/4 + (5*a^2*b*d^2)/
16 - (11*a*b^2*c*d)/16))/(a^3*c*d*((c + d*x)^(1/2) - c^(1/2))^5) + (((a + b*x)^(1/2) - a^(1/2))^4*((a^4*d^4)/3
2 - (7*b^4*c^4)/32 + (21*a^2*b^2*c^2*d^2)/32 + (a*b^3*c^3*d)/2 - (a^3*b*c*d^3)/2))/(a^(7/2)*c^(5/2)*d^2*((c +
d*x)^(1/2) - c^(1/2))^4))/(((a + b*x)^(1/2) - a^(1/2))^6/((c + d*x)^(1/2) - c^(1/2))^6 + (b^2*((a + b*x)^(1/2)
- a^(1/2))^2)/(d^2*((c + d*x)^(1/2) - c^(1/2))^2) + (((a + b*x)^(1/2) - a^(1/2))^4*(a^2*d^2 + b^2*c^2 + 4*a*b
*c*d))/(a*c*d^2*((c + d*x)^(1/2) - c^(1/2))^4) - ((2*b^2*c + 2*a*b*d)*((a + b*x)^(1/2) - a^(1/2))^3)/(a^(1/2)*
c^(1/2)*d^2*((c + d*x)^(1/2) - c^(1/2))^3) - ((2*a*d + 2*b*c)*((a + b*x)^(1/2) - a^(1/2))^5)/(a^(1/2)*c^(1/2)*
d*((c + d*x)^(1/2) - c^(1/2))^5)) + (d^2*((a + b*x)^(1/2) - a^(1/2))^2)/(32*a^(3/2)*c^(1/2)*((c + d*x)^(1/2) -
c^(1/2))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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