3.17.29 \(\int \frac {(a+b x) \sqrt {e+f x}}{c+d x} \, dx\)

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

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Rubi [A]  time = 0.08, 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} \begin {gather*} -\frac {2 \sqrt {e+f x} (b c-a d)}{d^2}+\frac {2 (b c-a d) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2}}+\frac {2 b (e+f x)^{3/2}}{3 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sqrt[e + f*x])/(c + d*x),x]

[Out]

(-2*(b*c - a*d)*Sqrt[e + f*x])/d^2 + (2*b*(e + f*x)^(3/2))/(3*d*f) + (2*(b*c - a*d)*Sqrt[d*e - c*f]*ArcTanh[(S
qrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(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 {e+f x}}{c+d x} \, dx &=\frac {2 b (e+f x)^{3/2}}{3 d f}+\frac {\left (2 \left (-\frac {3}{2} b c f+\frac {3 a d f}{2}\right )\right ) \int \frac {\sqrt {e+f x}}{c+d x} \, dx}{3 d f}\\ &=-\frac {2 (b c-a d) \sqrt {e+f x}}{d^2}+\frac {2 b (e+f x)^{3/2}}{3 d f}-\frac {((b c-a d) (d e-c f)) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d^2}\\ &=-\frac {2 (b c-a d) \sqrt {e+f x}}{d^2}+\frac {2 b (e+f x)^{3/2}}{3 d f}-\frac {(2 (b c-a d) (d e-c f)) \operatorname {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{d^2 f}\\ &=-\frac {2 (b c-a d) \sqrt {e+f x}}{d^2}+\frac {2 b (e+f x)^{3/2}}{3 d f}+\frac {2 (b c-a d) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{5/2}}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Sqrt[e + f*x])/(c + d*x),x]

[Out]

(2*Sqrt[e + f*x]*(-3*b*c*f + 3*a*d*f + b*d*(e + f*x)))/(3*d^2*f) + (2*(b*c - a*d)*Sqrt[d*e - c*f]*ArcTanh[(Sqr
t[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(5/2)

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IntegrateAlgebraic [A]  time = 0.16, size = 117, normalized size = 1.19 \begin {gather*} \frac {2 (a d-b c) \sqrt {c f-d e} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{5/2}}+\frac {2 \left (3 a d f \sqrt {e+f x}-3 b c f \sqrt {e+f x}+b d (e+f x)^{3/2}\right )}{3 d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)*Sqrt[e + f*x])/(c + d*x),x]

[Out]

(2*(-3*b*c*f*Sqrt[e + f*x] + 3*a*d*f*Sqrt[e + f*x] + b*d*(e + f*x)^(3/2)))/(3*d^2*f) + (2*(-(b*c) + a*d)*Sqrt[
-(d*e) + c*f]*ArcTan[(Sqrt[d]*Sqrt[-(d*e) + c*f]*Sqrt[e + f*x])/(d*e - c*f)])/d^(5/2)

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fricas [A]  time = 1.43, size = 211, normalized size = 2.15 \begin {gather*} \left [-\frac {3 \, {\left (b c - a d\right )} f \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) - 2 \, {\left (b d f x + b d e - 3 \, {\left (b c - a d\right )} f\right )} \sqrt {f x + e}}{3 \, d^{2} f}, \frac {2 \, {\left (3 \, {\left (b c - a d\right )} f \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) + {\left (b d f x + b d e - 3 \, {\left (b c - a d\right )} f\right )} \sqrt {f x + e}\right )}}{3 \, d^{2} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.21, size = 130, normalized size = 1.33 \begin {gather*} \frac {2 \, {\left (b c^{2} f - a c d f - b c d e + a d^{2} e\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{\sqrt {c d f - d^{2} e} d^{2}} + \frac {2 \, {\left ({\left (f x + e\right )}^{\frac {3}{2}} b d^{2} f^{2} - 3 \, \sqrt {f x + e} b c d f^{3} + 3 \, \sqrt {f x + e} a d^{2} f^{3}\right )}}{3 \, d^{3} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(f*x+e)^(1/2)/(d*x+c),x)

[Out]

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

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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)*(f*x+e)^(1/2)/(d*x+c),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(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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mupad [B]  time = 0.13, size = 107, normalized size = 1.09 \begin {gather*} \sqrt {e+f\,x}\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )+\frac {2\,b\,{\left (e+f\,x\right )}^{3/2}}{3\,d\,f}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}}{\sqrt {d\,e-c\,f}}\right )\,\left (a\,d-b\,c\right )\,\sqrt {d\,e-c\,f}}{d^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)^(1/2)*(a + b*x))/(c + d*x),x)

[Out]

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

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sympy [A]  time = 8.63, size = 94, normalized size = 0.96 \begin {gather*} \frac {2 \left (\frac {b \left (e + f x\right )^{\frac {3}{2}}}{3 d} + \frac {\sqrt {e + f x} \left (a d f - b c f\right )}{d^{2}} - \frac {f \left (a d - b c\right ) \left (c f - d e\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{3} \sqrt {\frac {c f - d e}{d}}}\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(f*x+e)**(1/2)/(d*x+c),x)

[Out]

2*(b*(e + f*x)**(3/2)/(3*d) + sqrt(e + f*x)*(a*d*f - b*c*f)/d**2 - f*(a*d - b*c)*(c*f - d*e)*atan(sqrt(e + f*x
)/sqrt((c*f - d*e)/d))/(d**3*sqrt((c*f - d*e)/d)))/f

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