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

Optimal. Leaf size=86 \[ -\frac {2 (b B-A c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c} \sqrt {c d-b e}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b \sqrt {d}} \]

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Rubi [A]  time = 0.18, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {826, 1166, 208} \begin {gather*} -\frac {2 (b B-A c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c} \sqrt {c d-b e}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d]) - (2*(b*B - A*c)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d -
b*e]])/(b*Sqrt[c]*Sqrt[c*d - b*e])

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 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 84, normalized size = 0.98 \begin {gather*} \frac {2 \left (\frac {(A c-b B) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{\sqrt {c} \sqrt {c d-b e}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-((A*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/Sqrt[d]) + ((-(b*B) + A*c)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d -
b*e]])/(Sqrt[c]*Sqrt[c*d - b*e])))/b

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IntegrateAlgebraic [A]  time = 0.16, size = 96, normalized size = 1.12 \begin {gather*} \frac {2 (A c-b B) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x} \sqrt {b e-c d}}{c d-b e}\right )}{b \sqrt {c} \sqrt {b e-c d}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(b*B) + A*c)*ArcTan[(Sqrt[c]*Sqrt[-(c*d) + b*e]*Sqrt[d + e*x])/(c*d - b*e)])/(b*Sqrt[c]*Sqrt[-(c*d) + b*e
]) - (2*A*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d])

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fricas [A]  time = 0.49, size = 489, normalized size = 5.69 \begin {gather*} \left [-\frac {\sqrt {c^{2} d - b c e} {\left (B b - A c\right )} d \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {c^{2} d - b c e} \sqrt {e x + d}}{c x + b}\right ) - {\left (A c^{2} d - A b c e\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right )}{b c^{2} d^{2} - b^{2} c d e}, \frac {2 \, \sqrt {-c^{2} d + b c e} {\left (B b - A c\right )} d \arctan \left (\frac {\sqrt {-c^{2} d + b c e} \sqrt {e x + d}}{c e x + c d}\right ) + {\left (A c^{2} d - A b c e\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right )}{b c^{2} d^{2} - b^{2} c d e}, -\frac {\sqrt {c^{2} d - b c e} {\left (B b - A c\right )} d \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {c^{2} d - b c e} \sqrt {e x + d}}{c x + b}\right ) - 2 \, {\left (A c^{2} d - A b c e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right )}{b c^{2} d^{2} - b^{2} c d e}, \frac {2 \, {\left (\sqrt {-c^{2} d + b c e} {\left (B b - A c\right )} d \arctan \left (\frac {\sqrt {-c^{2} d + b c e} \sqrt {e x + d}}{c e x + c d}\right ) + {\left (A c^{2} d - A b c e\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right )\right )}}{b c^{2} d^{2} - b^{2} c d e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(sqrt(c^2*d - b*c*e)*(B*b - A*c)*d*log((c*e*x + 2*c*d - b*e + 2*sqrt(c^2*d - b*c*e)*sqrt(e*x + d))/(c*x + b)
) - (A*c^2*d - A*b*c*e)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x))/(b*c^2*d^2 - b^2*c*d*e), (2*sqrt
(-c^2*d + b*c*e)*(B*b - A*c)*d*arctan(sqrt(-c^2*d + b*c*e)*sqrt(e*x + d)/(c*e*x + c*d)) + (A*c^2*d - A*b*c*e)*
sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x))/(b*c^2*d^2 - b^2*c*d*e), -(sqrt(c^2*d - b*c*e)*(B*b - A*
c)*d*log((c*e*x + 2*c*d - b*e + 2*sqrt(c^2*d - b*c*e)*sqrt(e*x + d))/(c*x + b)) - 2*(A*c^2*d - A*b*c*e)*sqrt(-
d)*arctan(sqrt(e*x + d)*sqrt(-d)/d))/(b*c^2*d^2 - b^2*c*d*e), 2*(sqrt(-c^2*d + b*c*e)*(B*b - A*c)*d*arctan(sqr
t(-c^2*d + b*c*e)*sqrt(e*x + d)/(c*e*x + c*d)) + (A*c^2*d - A*b*c*e)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d)
)/(b*c^2*d^2 - b^2*c*d*e)]

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giac [A]  time = 0.19, size = 79, normalized size = 0.92 \begin {gather*} \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b} + \frac {2 \, A \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 101, normalized size = 1.17 \begin {gather*} -\frac {2 A c \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}+\frac {2 B \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 A \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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mupad [B]  time = 1.92, size = 1130, normalized size = 13.14 \begin {gather*} -\frac {2\,A\,\mathrm {atanh}\left (\frac {16\,A^3\,c^2\,e^3\,\sqrt {d+e\,x}}{d^{3/2}\,\left (\frac {16\,A^3\,c^2\,e^3}{d}-32\,A^2\,B\,c^2\,e^2+16\,A\,B^2\,b\,c\,e^2\right )}-\frac {32\,A^2\,B\,c^2\,e^2\,\sqrt {d+e\,x}}{\sqrt {d}\,\left (\frac {16\,A^3\,c^2\,e^3}{d}-32\,A^2\,B\,c^2\,e^2+16\,A\,B^2\,b\,c\,e^2\right )}+\frac {16\,A\,B^2\,b\,c\,e^2\,\sqrt {d+e\,x}}{\sqrt {d}\,\left (\frac {16\,A^3\,c^2\,e^3}{d}-32\,A^2\,B\,c^2\,e^2+16\,A\,B^2\,b\,c\,e^2\right )}\right )}{b\,\sqrt {d}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\sqrt {d+e\,x}\,\left (16\,A^2\,c^3\,e^2-16\,A\,B\,b\,c^2\,e^2+8\,B^2\,b^2\,c\,e^2\right )+\frac {\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\left (8\,B\,b^2\,c^2\,d\,e^2-8\,A\,b^2\,c^2\,e^3+\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\sqrt {d+e\,x}}{b\,c^2\,d-b^2\,c\,e}\right )}{b\,c^2\,d-b^2\,c\,e}\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,1{}\mathrm {i}}{b\,c^2\,d-b^2\,c\,e}+\frac {\left (\sqrt {d+e\,x}\,\left (16\,A^2\,c^3\,e^2-16\,A\,B\,b\,c^2\,e^2+8\,B^2\,b^2\,c\,e^2\right )+\frac {\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\left (8\,A\,b^2\,c^2\,e^3-8\,B\,b^2\,c^2\,d\,e^2+\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\sqrt {d+e\,x}}{b\,c^2\,d-b^2\,c\,e}\right )}{b\,c^2\,d-b^2\,c\,e}\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,1{}\mathrm {i}}{b\,c^2\,d-b^2\,c\,e}}{\frac {\left (\sqrt {d+e\,x}\,\left (16\,A^2\,c^3\,e^2-16\,A\,B\,b\,c^2\,e^2+8\,B^2\,b^2\,c\,e^2\right )+\frac {\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\left (8\,B\,b^2\,c^2\,d\,e^2-8\,A\,b^2\,c^2\,e^3+\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\sqrt {d+e\,x}}{b\,c^2\,d-b^2\,c\,e}\right )}{b\,c^2\,d-b^2\,c\,e}\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}}{b\,c^2\,d-b^2\,c\,e}-\frac {\left (\sqrt {d+e\,x}\,\left (16\,A^2\,c^3\,e^2-16\,A\,B\,b\,c^2\,e^2+8\,B^2\,b^2\,c\,e^2\right )+\frac {\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\left (8\,A\,b^2\,c^2\,e^3-8\,B\,b^2\,c^2\,d\,e^2+\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\sqrt {d+e\,x}}{b\,c^2\,d-b^2\,c\,e}\right )}{b\,c^2\,d-b^2\,c\,e}\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}}{b\,c^2\,d-b^2\,c\,e}+16\,A^2\,B\,c^2\,e^2-16\,A\,B^2\,b\,c\,e^2}\right )\,\left (A\,c-B\,b\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,2{}\mathrm {i}}{b\,c^2\,d-b^2\,c\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*A*atanh((16*A^3*c^2*e^3*(d + e*x)^(1/2))/(d^(3/2)*((16*A^3*c^2*e^3)/d - 32*A^2*B*c^2*e^2 + 16*A*B^2*b*c*e
^2)) - (32*A^2*B*c^2*e^2*(d + e*x)^(1/2))/(d^(1/2)*((16*A^3*c^2*e^3)/d - 32*A^2*B*c^2*e^2 + 16*A*B^2*b*c*e^2))
+ (16*A*B^2*b*c*e^2*(d + e*x)^(1/2))/(d^(1/2)*((16*A^3*c^2*e^3)/d - 32*A^2*B*c^2*e^2 + 16*A*B^2*b*c*e^2))))/(
b*d^(1/2)) - (atan(((((d + e*x)^(1/2)*(16*A^2*c^3*e^2 + 8*B^2*b^2*c*e^2 - 16*A*B*b*c^2*e^2) + ((A*c - B*b)*(-c
*(b*e - c*d))^(1/2)*(8*B*b^2*c^2*d*e^2 - 8*A*b^2*c^2*e^3 + ((8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2)*(A*c - B*b)*(-c
*(b*e - c*d))^(1/2)*(d + e*x)^(1/2))/(b*c^2*d - b^2*c*e)))/(b*c^2*d - b^2*c*e))*(A*c - B*b)*(-c*(b*e - c*d))^(
1/2)*1i)/(b*c^2*d - b^2*c*e) + (((d + e*x)^(1/2)*(16*A^2*c^3*e^2 + 8*B^2*b^2*c*e^2 - 16*A*B*b*c^2*e^2) + ((A*c
- B*b)*(-c*(b*e - c*d))^(1/2)*(8*A*b^2*c^2*e^3 - 8*B*b^2*c^2*d*e^2 + ((8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2)*(A*c
- B*b)*(-c*(b*e - c*d))^(1/2)*(d + e*x)^(1/2))/(b*c^2*d - b^2*c*e)))/(b*c^2*d - b^2*c*e))*(A*c - B*b)*(-c*(b*
e - c*d))^(1/2)*1i)/(b*c^2*d - b^2*c*e))/((((d + e*x)^(1/2)*(16*A^2*c^3*e^2 + 8*B^2*b^2*c*e^2 - 16*A*B*b*c^2*e
^2) + ((A*c - B*b)*(-c*(b*e - c*d))^(1/2)*(8*B*b^2*c^2*d*e^2 - 8*A*b^2*c^2*e^3 + ((8*b^3*c^2*e^3 - 16*b^2*c^3*
d*e^2)*(A*c - B*b)*(-c*(b*e - c*d))^(1/2)*(d + e*x)^(1/2))/(b*c^2*d - b^2*c*e)))/(b*c^2*d - b^2*c*e))*(A*c - B
*b)*(-c*(b*e - c*d))^(1/2))/(b*c^2*d - b^2*c*e) - (((d + e*x)^(1/2)*(16*A^2*c^3*e^2 + 8*B^2*b^2*c*e^2 - 16*A*B
*b*c^2*e^2) + ((A*c - B*b)*(-c*(b*e - c*d))^(1/2)*(8*A*b^2*c^2*e^3 - 8*B*b^2*c^2*d*e^2 + ((8*b^3*c^2*e^3 - 16*
b^2*c^3*d*e^2)*(A*c - B*b)*(-c*(b*e - c*d))^(1/2)*(d + e*x)^(1/2))/(b*c^2*d - b^2*c*e)))/(b*c^2*d - b^2*c*e))*
(A*c - B*b)*(-c*(b*e - c*d))^(1/2))/(b*c^2*d - b^2*c*e) + 16*A^2*B*c^2*e^2 - 16*A*B^2*b*c*e^2))*(A*c - B*b)*(-
c*(b*e - c*d))^(1/2)*2i)/(b*c^2*d - b^2*c*e)

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sympy [A]  time = 69.75, size = 87, normalized size = 1.01 \begin {gather*} \frac {2 A \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{d}} \sqrt {d + e x}} \right )}}{b d \sqrt {- \frac {1}{d}}} - \frac {2 \left (- A c + B b\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {c}{b e - c d}} \sqrt {d + e x}} \right )}}{b \sqrt {\frac {c}{b e - c d}} \left (b e - c d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*atan(1/(sqrt(-1/d)*sqrt(d + e*x)))/(b*d*sqrt(-1/d)) - 2*(-A*c + B*b)*atan(1/(sqrt(c/(b*e - c*d))*sqrt(d +
e*x)))/(b*sqrt(c/(b*e - c*d))*(b*e - c*d))

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