3.2222 \(\int \frac{A+B x}{\sqrt{a+b x} \sqrt{d+e x}} \, dx\)

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]*Sqrt[d + e*x])])/(b^(3/2)*e^(3/2))

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Rubi [A]  time = 0.163738, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125 \[ \frac{(2 A b e-B (a e+b d)) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{3/2} e^{3/2}}+\frac{B \sqrt{a+b x} \sqrt{d+e x}}{b e} \]

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]*Sqrt[d + e*x])])/(b^(3/2)*e^(3/2))

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Rubi in Sympy [A]  time = 12.1138, size = 76, normalized size = 0.9 \[ \frac{B \sqrt{a + b x} \sqrt{d + e x}}{b e} - \frac{2 \left (- A b e + \frac{B \left (a e + b d\right )}{2}\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{d + e x}}{\sqrt{e} \sqrt{a + b x}} \right )}}{b^{\frac{3}{2}} e^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)/(b*x+a)**(1/2)/(e*x+d)**(1/2),x)

[Out]

B*sqrt(a + b*x)*sqrt(d + e*x)/(b*e) - 2*(-A*b*e + B*(a*e + b*d)/2)*atanh(sqrt(b)
*sqrt(d + e*x)/(sqrt(e)*sqrt(a + b*x)))/(b**(3/2)*e**(3/2))

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Mathematica [A]  time = 0.132936, size = 100, normalized size = 1.19 \[ \frac{(-a B e+2 A b e-b B d) \log \left (2 \sqrt{b} \sqrt{e} \sqrt{a+b x} \sqrt{d+e x}+a e+b d+2 b e x\right )}{2 b^{3/2} e^{3/2}}+\frac{B \sqrt{a+b x} \sqrt{d+e x}}{b e} \]

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)*Log[b*d +
a*e + 2*b*e*x + 2*Sqrt[b]*Sqrt[e]*Sqrt[a + b*x]*Sqrt[d + e*x]])/(2*b^(3/2)*e^(3/
2))

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Maple [B]  time = 0.029, size = 198, normalized size = 2.4 \[{\frac{1}{2\,be} \left ( 2\,A\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) be-B\ln \left ({\frac{1}{2} \left ( 2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd \right ){\frac{1}{\sqrt{be}}}} \right ) ae-B\ln \left ({\frac{1}{2} \left ( 2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd \right ){\frac{1}{\sqrt{be}}}} \right ) bd+2\,B\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be} \right ) \sqrt{bx+a}\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/(sqrt(b*x + a)*sqrt(e*x + d)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.446397, size = 1, normalized size = 0.01 \[ \left [\frac{4 \, \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} B -{\left (B b d +{\left (B a - 2 \, A b\right )} e\right )} \log \left (4 \,{\left (2 \, b^{2} e^{2} x + b^{2} d e + a b e^{2}\right )} \sqrt{b x + a} \sqrt{e x + d} +{\left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right )} \sqrt{b e}\right )}{4 \, \sqrt{b e} b e}, \frac{2 \, \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d} B -{\left (B b d +{\left (B a - 2 \, A b\right )} e\right )} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e}}{2 \, \sqrt{b x + a} \sqrt{e x + d} b e}\right )}{2 \, \sqrt{-b e} b e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/(sqrt(b*x + a)*sqrt(e*x + d)),x, algorithm="fricas")

[Out]

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

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

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/XCAS [A]  time = 0.230052, size = 143, normalized size = 1.7 \[ \frac{{\left (\frac{{\left (B b d + B a e - 2 \, A b e\right )} e^{\left (-\frac{3}{2}\right )}{\rm ln}\left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac{3}{2}}} + \frac{\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a} B e^{\left (-1\right )}}{b^{2}}\right )} b}{{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)/(sqrt(b*x + a)*sqrt(e*x + d)),x, algorithm="giac")

[Out]

((B*b*d + B*a*e - 2*A*b*e)*e^(-3/2)*ln(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)