∫ √ ____ c + dx√______

Optimal. Leaf size=72 \[ \frac {\sqrt {a+b x} \sqrt {c+d x}}{b}+\frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}} \]

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

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Rubi [A]
time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {52, 65, 223, 212} \begin {gather*} \frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b} \end {gather*}

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

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]

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,

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]

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,

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

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Mathematica [A]
time = 0.21, size = 74, normalized size = 1.03 \begin {gather*} \frac {b \sqrt {a+b x} \sqrt {c+d x}+\sqrt {\frac {b}{d}} (-b c+a d) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{b^2} \end {gather*}

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

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method	result	size

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

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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

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Fricas [A]
time = 0.67, size = 236, normalized size = 3.28 \begin {gather*} \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^{2} d}, \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^{2} d}\right ] \end {gather*}

[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^2*d), 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^2*d)]

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

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Giac [A]
time = 0.99, size = 93, normalized size = 1.29 \begin {gather*} -\frac {{\left (\frac {{\left (b^{2} c - a b 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}} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}\right )} {\left | b \right |}}{b^{3}} \end {gather*}

-((b^2*c - a*b*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) - sqrt(b^

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Mupad [B]
time = 4.01, size = 260, normalized size = 3.61 \begin {gather*} \frac {\frac {\left (2\,a\,d+2\,b\,c\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{d^2\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}+\frac {\left (2\,a\,d+2\,b\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{b\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {8\,\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d\,{\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 )}{b^{3/2}\,\sqrt {d}} \end {gather*}

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

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

d*((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^(3/2)*d^(1/2))

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3.15.64 \(\int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx\) [1464]