\(\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx\) [579]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {52, 65, 223, 212} \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}} \]

[In]

Int[Sqrt[a + b*x]/Sqrt[c + d*x],x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 212

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

Rule 223

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}+\frac {(-b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b} d^{3/2}} \]

[In]

Integrate[Sqrt[a + b*x]/Sqrt[c + d*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.47

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{d}-\frac {\left (-a d +b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 d \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}}\)	\(107\)

[In]

int((b*x+a)^(1/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(b*x+a)^(1/2)*(d*x+c)^(1/2)/d-1/2*(-a*d+b*c)/d*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d
+1/2*b*c+b*d*x)/(b*d)^(1/2)+(b*d*x^2+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.22 \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\left [\frac {4 \, \sqrt {b x + a} \sqrt {d x + c} b d - {\left (b c - a d\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right )}{4 \, b d^{2}}, \frac {2 \, \sqrt {b x + a} \sqrt {d x + c} b d + {\left (b c - a d\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right )}{2 \, b d^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\int \frac {\sqrt {a + b x}}{\sqrt {c + d x}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\frac {b {\left (\frac {{\left (b c - a d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d} + \frac {\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}}{b d}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

b*((b*c - a*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d) + sqrt(b
^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)/(b*d))/abs(b)

Mupad [B] (verification not implemented)

Time = 4.27 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.58 \[ \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx=\frac {\frac {\left (2\,a\,d+2\,b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}+\frac {\left (2\,c\,b^2+2\,a\,d\,b\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{d^3\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {8\,\sqrt {a}\,b\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}+\frac {b^2}{d^2}-\frac {2\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}}+\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (a\,d-b\,c\right )}{\sqrt {b}\,d^{3/2}} \]

[In]

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

[Out]

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