3.23.39 \(\int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx\) [2239]

Optimal. Leaf size=84 \[ \frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}+\frac {(2 A b e-B (b d+a e)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{3/2} e^{3/2}} \]

[Out]

(2*A*b*e-B*(a*e+b*d))*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2))/b^(3/2)/e^(3/2)+B*(b*x+a)^(1/2)*(e*
x+d)^(1/2)/b/e

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Rubi [A]
time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {81, 65, 223, 212} \begin {gather*} \frac {(2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{3/2} e^{3/2}}+\frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(B*Sqrt[a + b*x]*Sqrt[d + e*x])/(b*e) + ((2*A*b*e - B*(b*d + a*e))*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sq
rt[d + e*x])])/(b^(3/2)*e^(3/2))

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

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Mathematica [A]
time = 0.19, size = 83, normalized size = 0.99 \begin {gather*} \frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}-\frac {(b B d-2 A b e+a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{3/2} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(B*Sqrt[a + b*x]*Sqrt[d + e*x])/(b*e) - ((b*B*d - 2*A*b*e + a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sq
rt[a + b*x])])/(b^(3/2)*e^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(197\) vs. \(2(68)=136\).
time = 0.10, size = 198, normalized size = 2.36

method	result	size
default	\(\frac {\left (2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b e -B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a e -B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b d +2 B \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right ) \sqrt {b x +a}\, \sqrt {e x +d}}{2 e \sqrt {b e}\, b \sqrt {\left (b x +a \right ) \left (e x +d \right )}}\)	\(198\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b*e-B*ln(1/2*(2*b*e*x+2*(
(b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*e-B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)
^(1/2)+a*e+b*d)/(b*e)^(1/2))*b*d+2*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))*(b*x+a)^(1/2)*(e*x+d)^(1/2)/e/(b*e)^
(1/2)/b/((b*x+a)*(e*x+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*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(1/2)/(e*x+d)^(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 0.82, size = 257, normalized size = 3.06 \begin {gather*} \left [\frac {{\left (4 \, \sqrt {b x + a} \sqrt {x e + d} B b e - {\left (B b d + {\left (B a - 2 \, A b\right )} e\right )} \sqrt {b} e^{\frac {1}{2}} \log \left (b^{2} d^{2} + 4 \, {\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\frac {1}{2}} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right )\right )} e^{\left (-2\right )}}{4 \, b^{2}}, \frac {{\left (2 \, \sqrt {b x + a} \sqrt {x e + d} B b e + {\left (B b d + {\left (B a - 2 \, A b\right )} e\right )} \sqrt {-b e} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {-b e} \sqrt {x e + d}}{2 \, {\left ({\left (b^{2} x^{2} + a b x\right )} e^{2} + {\left (b^{2} d x + a b d\right )} e\right )}}\right )\right )} e^{\left (-2\right )}}{2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(b*x + a)*sqrt(x*e + d)*B*b*e - (B*b*d + (B*a - 2*A*b)*e)*sqrt(b)*e^(1/2)*log(b^2*d^2 + 4*(b*d + (
2*b*x + a)*e)*sqrt(b*x + a)*sqrt(x*e + d)*sqrt(b)*e^(1/2) + (8*b^2*x^2 + 8*a*b*x + a^2)*e^2 + 2*(4*b^2*d*x + 3
*a*b*d)*e))*e^(-2)/b^2, 1/2*(2*sqrt(b*x + a)*sqrt(x*e + d)*B*b*e + (B*b*d + (B*a - 2*A*b)*e)*sqrt(-b*e)*arctan
(1/2*(b*d + (2*b*x + a)*e)*sqrt(b*x + a)*sqrt(-b*e)*sqrt(x*e + d)/((b^2*x^2 + a*b*x)*e^2 + (b^2*d*x + a*b*d)*e
)))*e^(-2)/b^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)**(1/2)/(e*x+d)**(1/2),x)

[Out]

Integral((A + B*x)/(sqrt(a + b*x)*sqrt(d + e*x)), x)

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Giac [A]
time = 1.14, size = 106, normalized size = 1.26 \begin {gather*} \frac {{\left (\frac {{\left (B b d + B a e - 2 \, A b e\right )} e^{\left (-\frac {3}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} B e^{\left (-1\right )}}{b^{2}}\right )} b}{{\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*b*d + B*a*e - 2*A*b*e)*e^(-3/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*
e)))/b^(3/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*B*e^(-1)/b^2)*b/abs(b)

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Mupad [B]
time = 5.42, size = 311, normalized size = 3.70 \begin {gather*} \frac {\frac {\left (2\,B\,a\,e+2\,B\,b\,d\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{e^3\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}+\frac {\left (2\,B\,a\,e+2\,B\,b\,d\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{b\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}-\frac {8\,B\,\sqrt {a}\,\sqrt {d}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}+\frac {b^2}{e^2}-\frac {2\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}-\frac {4\,A\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {-b\,e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {-b\,e}}-\frac {2\,B\,\mathrm {atanh}\left (\frac {\sqrt {e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )\,\left (a\,e+b\,d\right )}{b^{3/2}\,e^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)^(1/2)*(d + e*x)^(1/2)),x)

[Out]

(((2*B*a*e + 2*B*b*d)*((a + b*x)^(1/2) - a^(1/2)))/(e^3*((d + e*x)^(1/2) - d^(1/2))) + ((2*B*a*e + 2*B*b*d)*((
a + b*x)^(1/2) - a^(1/2))^3)/(b*e^2*((d + e*x)^(1/2) - d^(1/2))^3) - (8*B*a^(1/2)*d^(1/2)*((a + b*x)^(1/2) - a
^(1/2))^2)/(e^2*((d + e*x)^(1/2) - d^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2))^4/((d + e*x)^(1/2) - d^(1/2))^4 +
b^2/e^2 - (2*b*((a + b*x)^(1/2) - a^(1/2))^2)/(e*((d + e*x)^(1/2) - d^(1/2))^2)) - (4*A*atan((b*((d + e*x)^(1
/2) - d^(1/2)))/((-b*e)^(1/2)*((a + b*x)^(1/2) - a^(1/2)))))/(-b*e)^(1/2) - (2*B*atanh((e^(1/2)*((a + b*x)^(1/
2) - a^(1/2)))/(b^(1/2)*((d + e*x)^(1/2) - d^(1/2))))*(a*e + b*d))/(b^(3/2)*e^(3/2))

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