3.17.41 \(\int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=173 \[ \frac {2 (a+b x) \sqrt {d+e x} (A b-a B)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (A b-a B) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.10, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 80, 50, 63, 208} \begin {gather*} \frac {2 (a+b x) \sqrt {d+e x} (A b-a B)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (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} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[d + e*x])/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(A+B x) \sqrt {d+e x}}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (\frac {3}{2} A b^2 e-\frac {3}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{3 b^2 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (a+b x) \sqrt {d+e x}}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right ) \left (\frac {3}{2} A b^2 e-\frac {3}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{3 b^4 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (a+b x) \sqrt {d+e x}}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 \left (b^2 d-a b e\right ) \left (\frac {3}{2} A b^2 e-\frac {3}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{3 b^4 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (a+b x) \sqrt {d+e x}}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{3/2}}{3 b e \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) \sqrt {b d-a e} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 114, normalized size = 0.66 \begin {gather*} \frac {2 (a+b x) \left (\sqrt {b} \sqrt {d+e x} (-3 a B e+3 A b e+b B (d+e x))+3 e (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 )\right )}{3 b^{5/2} e \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[d + e*x])/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 28.93, size = 136, normalized size = 0.79 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {2 (A b-a B) \sqrt {a e-b d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{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}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*Sqrt[d + e*x])/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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fricas [A]  time = 0.43, size = 211, normalized size = 1.22 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/((b*x+a)^2)^(1/2),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 = 0.24, size = 168, normalized size = 0.97 \begin {gather*} -\frac {2 \, {\left (B a b d \mathrm {sgn}\left (b x + a\right ) - A b^{2} d \mathrm {sgn}\left (b x + a\right ) - B a^{2} e \mathrm {sgn}\left (b x + a\right ) + A a b e \mathrm {sgn}\left (b x + a\right )\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} \mathrm {sgn}\left (b x + a\right ) - 3 \, \sqrt {x e + d} B a b e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, \sqrt {x e + d} A b^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 226, normalized size = 1.31 \begin {gather*} \frac {2 \left (b x +a \right ) \left (-3 A a b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 A \,b^{2} d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 B \,a^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-3 B a b d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, A b e -3 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, B a e +\left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, B b \right )}{3 \sqrt {\left (b x +a \right )^{2}}\, \sqrt {\left (a e -b d \right ) b}\, b^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x + A\right )} \sqrt {e x + d}}{\sqrt {{\left (b x + a\right )}^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {d+e\,x}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \sqrt {d + e x}}{\sqrt {\left (a + b x\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*sqrt(d + e*x)/sqrt((a + b*x)**2), x)

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