3.11.58 \(\int \frac {A+B x}{(d+e x) (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=141 \[ -\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (c d-b e)}-\frac {e (B d-A e) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {822, 12, 724, 206} \begin {gather*} -\frac {2 (c x (2 A c d-b (A e+B d))+A b (c d-b e))}{b^2 d \sqrt {b x+c x^2} (c d-b e)}-\frac {e (B d-A e) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)*(b*x + c*x^2)^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

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

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 144, normalized size = 1.02 \begin {gather*} \frac {2 \left (\sqrt {d} \sqrt {c d-b e} \left (A \left (b^2 e-b c d+b c e x-2 c^2 d x\right )+b B c d x\right )+b^2 e \sqrt {x} \sqrt {b+c x} (A e-B d) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )\right )}{b^2 d^{3/2} \sqrt {x (b+c x)} (c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)*(b*x + c*x^2)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.62, size = 156, normalized size = 1.11 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (A b^2 e-A b c d+A b c e x-2 A c^2 d x+b B c d x\right )}{b^2 d x (b+c x) (b e-c d)}-\frac {2 \left (B d e-A e^2\right ) \tanh ^{-1}\left (\frac {-e \sqrt {b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {d} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/((d + e*x)*(b*x + c*x^2)^(3/2)),x]

[Out]

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

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fricas [B]  time = 0.43, size = 543, normalized size = 3.85 \begin {gather*} \left [\frac {\sqrt {c d^{2} - b d e} {\left ({\left (B b^{2} c d e - A b^{2} c e^{2}\right )} x^{2} + {\left (B b^{3} d e - A b^{3} e^{2}\right )} x\right )} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) - 2 \, {\left (A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} + {\left (A b^{2} c d e^{2} - {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{3} + {\left (B b^{2} c - 3 \, A b c^{2}\right )} d^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}, -\frac {2 \, {\left (\sqrt {-c d^{2} + b d e} {\left ({\left (B b^{2} c d e - A b^{2} c e^{2}\right )} x^{2} + {\left (B b^{3} d e - A b^{3} e^{2}\right )} x\right )} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) + {\left (A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} + {\left (A b^{2} c d e^{2} - {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{3} + {\left (B b^{2} c - 3 \, A b c^{2}\right )} d^{2} e\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e + b^{4} c d^{2} e^{2}\right )} x^{2} + {\left (b^{3} c^{2} d^{4} - 2 \, b^{4} c d^{3} e + b^{5} d^{2} e^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[(sqrt(c*d^2 - b*d*e)*((B*b^2*c*d*e - A*b^2*c*e^2)*x^2 + (B*b^3*d*e - A*b^3*e^2)*x)*log((b*d + (2*c*d - b*e)*x
- 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(A*b*c^2*d^3 - 2*A*b^2*c*d^2*e + A*b^3*d*e^2 + (A*b
^2*c*d*e^2 - (B*b*c^2 - 2*A*c^3)*d^3 + (B*b^2*c - 3*A*b*c^2)*d^2*e)*x)*sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^
3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2*d^4 - 2*b^4*c*d^3*e + b^5*d^2*e^2)*x), -2*(sqrt(-c*d^2 + b*d*e)*((
B*b^2*c*d*e - A*b^2*c*e^2)*x^2 + (B*b^3*d*e - A*b^3*e^2)*x)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c
*d - b*e)*x)) + (A*b*c^2*d^3 - 2*A*b^2*c*d^2*e + A*b^3*d*e^2 + (A*b^2*c*d*e^2 - (B*b*c^2 - 2*A*c^3)*d^3 + (B*b
^2*c - 3*A*b*c^2)*d^2*e)*x)*sqrt(c*x^2 + b*x))/((b^2*c^3*d^4 - 2*b^3*c^2*d^3*e + b^4*c*d^2*e^2)*x^2 + (b^3*c^2
*d^4 - 2*b^4*c*d^3*e + b^5*d^2*e^2)*x)]

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giac [A]  time = 0.28, size = 187, normalized size = 1.33 \begin {gather*} \frac {2 \, {\left (\frac {{\left (B b c d^{2} - 2 \, A c^{2} d^{2} + A b c d e\right )} x}{b^{2} c d^{3} - b^{3} d^{2} e} - \frac {A b c d^{2} - A b^{2} d e}{b^{2} c d^{3} - b^{3} d^{2} e}\right )}}{\sqrt {c x^{2} + b x}} + \frac {2 \, {\left (B d e - A e^{2}\right )} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right )}{{\left (c d^{2} - b d e\right )} \sqrt {-c d^{2} + b d e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 837, normalized size = 5.94 \begin {gather*} -\frac {2 A c e x}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b d}+\frac {4 A \,c^{2} x}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b^{2}}+\frac {A e \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, d}+\frac {2 B c x}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b}-\frac {4 B \,c^{2} d x}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b^{2} e}-\frac {B \ln \left (\frac {-\frac {2 \left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right ) \sqrt {-\frac {\left (b e -c d \right ) d}{e^{2}}}}+\frac {2 A c}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b}-\frac {2 A e}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, d}-\frac {2 B c d}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}\, b e}+\frac {2 B}{\left (b e -c d \right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {\left (b e -c d \right ) d}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}-\frac {2 \left (2 c x +b \right ) B}{\sqrt {c \,x^{2}+b x}\, b^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)/(c*x^2+b*x)^(3/2),x)

[Out]

-2*B/e*(2*c*x+b)/b^2/(c*x^2+b*x)^(1/2)-2*e/(b*e-c*d)/d/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/
2)*A+2/(b*e-c*d)/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*B-2*e/(b*e-c*d)/d/b/((x+d/e)^2*c-(b
*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*c*A+2/(b*e-c*d)/b/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e
)/e)^(1/2)*x*c*B+4/(b*e-c*d)/b^2/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*c^2*A-4/e/(b*e-c*
d)/b^2/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*x*c^2*B*d+2/(b*e-c*d)/b/((x+d/e)^2*c-(b*e-c*d
)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2)*c*A-2/e/(b*e-c*d)/b/((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(
1/2)*c*B*d+e/(b*e-c*d)/d/(-(b*e-c*d)*d/e^2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d
/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*A-1/(b*e-c*d)/(-(b*e-c*d)*d/e^
2)^(1/2)*ln((-2*(b*e-c*d)*d/e^2+(b*e-2*c*d)*(x+d/e)/e+2*(-(b*e-c*d)*d/e^2)^(1/2)*((x+d/e)^2*c-(b*e-c*d)*d/e^2+
(b*e-2*c*d)*(x+d/e)/e)^(1/2))/(x+d/e))*B

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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)/(c*x^2+b*x)^(3/2),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-(2*c*d)/e^2)^2>0)', see `
assume?` for more details)Is (b/e-(2*c*d)/e^2)^2    -(4*c       *((c*d^2)/e^2        -(b*d)/e))     /e^2 zero
or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((b*x + c*x^2)^(3/2)*(d + e*x)),x)

[Out]

int((A + B*x)/((b*x + c*x^2)^(3/2)*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral((A + B*x)/((x*(b + c*x))**(3/2)*(d + e*x)), x)

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