3.1745 \(\int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx\)

Optimal. Leaf size=98 \[ -\frac {2 (A b-a B) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}+\frac {2 \sqrt {d+e x} (A b-a B)}{b^2}+\frac {2 B (d+e x)^{3/2}}{3 b e} \]

[Out]

2/3*B*(e*x+d)^(3/2)/b/e-2*(A*b-B*a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*(-a*e+b*d)^(1/2)/b^(5/2)+2
*(A*b-B*a)*(e*x+d)^(1/2)/b^2

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \[ \frac {2 \sqrt {d+e x} (A b-a B)}{b^2}-\frac {2 (A b-a B) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}+\frac {2 B (d+e x)^{3/2}}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*Sqrt[d + e*x])/b^2 + (2*B*(d + e*x)^(3/2))/(3*b*e) - (2*(A*b - a*B)*Sqrt[b*d - a*e]*ArcTanh[(Sq
rt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(5/2)

Rule 50

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 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 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 208

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

Rubi steps

\begin {align*} \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx &=\frac {2 B (d+e x)^{3/2}}{3 b e}+\frac {\left (2 \left (\frac {3 A b e}{2}-\frac {3 a B e}{2}\right )\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{3 b e}\\ &=\frac {2 (A b-a B) \sqrt {d+e x}}{b^2}+\frac {2 B (d+e x)^{3/2}}{3 b e}+\frac {((A b-a B) (b d-a e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{b^2}\\ &=\frac {2 (A b-a B) \sqrt {d+e x}}{b^2}+\frac {2 B (d+e x)^{3/2}}{3 b e}+\frac {(2 (A b-a B) (b d-a e)) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b^2 e}\\ &=\frac {2 (A b-a B) \sqrt {d+e x}}{b^2}+\frac {2 B (d+e x)^{3/2}}{3 b e}-\frac {2 (A b-a B) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 94, normalized size = 0.96 \[ \frac {2 (a B-A b) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}}+\frac {2 \sqrt {d+e x} (-3 a B e+3 A b e+b B (d+e x))}{3 b^2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(3*A*b*e - 3*a*B*e + b*B*(d + e*x)))/(3*b^2*e) + (2*(-(A*b) + a*B)*Sqrt[b*d - a*e]*ArcTanh[(S
qrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(5/2)

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fricas [A]  time = 0.97, size = 211, normalized size = 2.15 \[ \left [-\frac {3 \, {\left (B a - A b\right )} e \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (B b e x + B b d - 3 \, {\left (B a - A b\right )} e\right )} \sqrt {e x + d}}{3 \, b^{2} e}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} e \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (B b e x + B b d - 3 \, {\left (B a - A b\right )} e\right )} \sqrt {e x + d}\right )}}{3 \, b^{2} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.23, size = 126, normalized size = 1.29 \[ -\frac {2 \, {\left (B a b d - A b^{2} d - B a^{2} e + A a b e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} B b^{2} e^{2} - 3 \, \sqrt {x e + d} B a b e^{3} + 3 \, \sqrt {x e + d} A b^{2} e^{3}\right )} e^{\left (-3\right )}}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 211, normalized size = 2.15 \[ -\frac {2 A a e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 A d \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}+\frac {2 B \,a^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {2 B a d \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 \sqrt {e x +d}\, A}{b}-\frac {2 \sqrt {e x +d}\, B a}{b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B}{3 b e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a),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*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 1.30, size = 107, normalized size = 1.09 \[ \left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\sqrt {d+e\,x}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {b\,d-a\,e}}\right )\,\left (A\,b-B\,a\right )\,\sqrt {b\,d-a\,e}}{b^{5/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{3/2}}{3\,b\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(d + e*x)^(1/2) - (2*atanh((b^(1/2)*(d + e*x)^(1/2))
/(b*d - a*e)^(1/2))*(A*b - B*a)*(b*d - a*e)^(1/2))/b^(5/2) + (2*B*(d + e*x)^(3/2))/(3*b*e)

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sympy [A]  time = 9.07, size = 94, normalized size = 0.96 \[ \frac {2 \left (\frac {B \left (d + e x\right )^{\frac {3}{2}}}{3 b} + \frac {\sqrt {d + e x} \left (A b e - B a e\right )}{b^{2}} + \frac {e \left (- A b + B a\right ) \left (a e - b d\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{3} \sqrt {\frac {a e - b d}{b}}}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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