3.2239 \(\int \frac{A+B x}{\sqrt{a+b x} \sqrt{d+e x}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0598965, antiderivative size = 84, normalized size of antiderivative = 1., 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 = {80, 63, 217, 206} \[ \frac{(2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{3/2} e^{3/2}}+\frac{B \sqrt{a+b x} \sqrt{d+e x}}{b e} \]

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 80

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 63

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 217

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]

Rule 206

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

Mathematica [A]  time = 0.189946, size = 118, normalized size = 1.4 \[ \frac{b B \sqrt{e} \sqrt{a+b x} (d+e x)-\sqrt{b d-a e} \sqrt{\frac{b (d+e x)}{b d-a e}} (a B e-2 A b e+b B d) \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{b^2 e^{3/2} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.019, size = 198, normalized size = 2.4 \begin{align*}{\frac{1}{2\,be} \left ( 2\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) be-B\ln \left ({\frac{1}{2} \left ( 2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd \right ){\frac{1}{\sqrt{be}}}} \right ) ae-B\ln \left ({\frac{1}{2} \left ( 2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd \right ){\frac{1}{\sqrt{be}}}} \right ) bd+2\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be} \right ) \sqrt{bx+a}\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \end{align*}

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)

[Out]

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

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

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

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Fricas [A]  time = 1.3878, size = 606, normalized size = 7.21 \begin{align*} \left [\frac{4 \, \sqrt{b x + a} \sqrt{e x + d} B b e -{\left (B b d +{\left (B a - 2 \, A b\right )} e\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right )}{4 \, b^{2} e^{2}}, \frac{2 \, \sqrt{b x + a} \sqrt{e x + d} B b e +{\left (B b d +{\left (B a - 2 \, A b\right )} e\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right )}{2 \, b^{2} e^{2}}\right ] \end{align*}

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

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

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 = 1.97696, size = 143, normalized size = 1.7 \begin{align*} \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{align*}

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)