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

Optimal. Leaf size=122 \[ \frac {(b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 c x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 a c x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ \frac {(b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 c x^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 a c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.09, size = 99, normalized size = 0.81 \[ \frac {(b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{3/2} c^{3/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (2 a c+a d x+b c x)}{4 a c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 326, normalized size = 2.67 \[ \left [\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a c} x^{2} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, a^{2} c^{2} x^{2}}, -\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, a^{2} c^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 4.14, size = 1081, normalized size = 8.86 \[ \frac {\frac {{\left (\sqrt {b d} b^{3} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b^{2} c d {\left | b \right |} + \sqrt {b d} a^{2} b d^{2} {\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} a b c} - \frac {2 \, {\left (\sqrt {b d} b^{9} c^{5} {\left | b \right |} - 3 \, \sqrt {b d} a b^{8} c^{4} d {\left | b \right |} + 2 \, \sqrt {b d} a^{2} b^{7} c^{3} d^{2} {\left | b \right |} + 2 \, \sqrt {b d} a^{3} b^{6} c^{2} d^{3} {\left | b \right |} - 3 \, \sqrt {b d} a^{4} b^{5} c d^{4} {\left | b \right |} + \sqrt {b d} a^{5} b^{4} d^{5} {\left | b \right |} - 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^{7} c^{4} {\left | b \right |} + 4 \, \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^{6} c^{3} d {\left | b \right |} - 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 )}^{2} a^{2} b^{5} c^{2} d^{2} {\left | b \right |} + 4 \, \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^{4} c d^{3} {\left | b \right |} - 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^{3} d^{4} {\left | b \right |} + 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^{5} c^{3} {\left | b \right |} + 5 \, \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^{4} c^{2} d {\left | b \right |} + 5 \, \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^{3} c d^{2} {\left | b \right |} + 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^{2} d^{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 )}^{6} b^{3} c^{2} {\left | b \right |} - 6 \, \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^{2} c d {\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 )}^{6} a^{2} b d^{2} {\left | b \right |}\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 c}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((sqrt(b*d)*b^3*c^2*abs(b) - 2*sqrt(b*d)*a*b^2*c*d*abs(b) + sqrt(b*d)*a^2*b*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*c) - 2*(sqrt(b*d)*b^9*c^5*abs(b) - 3*sqrt(b*d)*a*b^8*c^4*d*abs(b) + 2*sqrt(b*d)*a^2*b^7*c^3*d^2*abs(b)
+ 2*sqrt(b*d)*a^3*b^6*c^2*d^3*abs(b) - 3*sqrt(b*d)*a^4*b^5*c*d^4*abs(b) + sqrt(b*d)*a^5*b^4*d^5*abs(b) - 3*sq
rt(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^4*abs(b) + 4*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^3*d*abs(b) - 2*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^2*d^2*abs(b) + 4*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*d^3*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*a^4*b^3*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^3*abs(b) + 5*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^2*d*abs(b) + 5*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*d^2*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*a^3*b^2*d^3*ab
s(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^2*abs(b) - 6*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^2*c*d*abs(b) - sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*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*c))/b

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maple [B]  time = 0.02, size = 305, normalized size = 2.50 \[ \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (a^{2} d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-2 a b c d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+b^{2} c^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a d x -2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b c x -4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a c \right )}{8 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {a c}\, a c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c*(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*d^2-2*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*a*b*c*d+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^2-2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*x*a*d-2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b*c-4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a*c*(a*c)^(
1/2))/(b*d*x^2+a*d*x+b*c*x+a*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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 19.75, size = 869, normalized size = 7.12 \[ \frac {\frac {b^4}{32\,\sqrt {a}\,\sqrt {c}\,d^2}-\frac {\left (\frac {c\,b^2}{16}+\frac {a\,d\,b}{16}\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}{a\,c\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2\,\left (\frac {5\,a^2\,b^2\,d^2}{32}+\frac {3\,a\,b^3\,c\,d}{8}+\frac {5\,b^4\,c^2}{32}\right )}{a^{3/2}\,c^{3/2}\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^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 {3\,a\,b^3\,c^2\,d}{8}+\frac {b^4\,c^3}{16}\right )}{a^2\,c^2\,d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {\left (\frac {c\,b^4}{8}+\frac {a\,d\,b^3}{8}\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{a\,c\,d^2\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (-\frac {a^4\,d^4}{32}+\frac {a^3\,b\,c\,d^3}{4}+\frac {3\,a^2\,b^2\,c^2\,d^2}{32}+\frac {a\,b^3\,c^3\,d}{4}-\frac {b^4\,c^4}{32}\right )}{a^{5/2}\,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 {\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 (\sqrt {a}\,b^2\,c^{5/2}+a^{5/2}\,\sqrt {c}\,d^2-2\,a^{3/2}\,b\,c^{3/2}\,d\right )}{8\,a^2\,c^2}+\frac {\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+d\,x}-\sqrt {c}}\right )\,\left (\sqrt {a}\,b^2\,c^{5/2}+a^{5/2}\,\sqrt {c}\,d^2-2\,a^{3/2}\,b\,c^{3/2}\,d\right )}{8\,a^2\,c^2}+\frac {d^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{32\,\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}-\frac {d\,\left (a\,d+b\,c\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{16\,a\,c\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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