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

Optimal. Leaf size=140 \[ -\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}+\frac {a B e-3 A b e+2 b B d}{b \sqrt {d+e x} (b d-a e)^2}-\frac {(a B e-3 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{5/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 78, 51, 63, 208} \begin {gather*} -\frac {A b-a B}{b (a+b x) \sqrt {d+e x} (b d-a e)}+\frac {a B e-3 A b e+2 b B d}{b \sqrt {d+e x} (b d-a e)^2}-\frac {(a B e-3 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [C]  time = 0.04, size = 95, normalized size = 0.68 \begin {gather*} \frac {(a+b x) (a B e-3 A b e+2 b B d) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )-(A b-a B) (b d-a e)}{b (a+b x) \sqrt {d+e x} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((A*b - a*B)*(b*d - a*e)) + (2*b*B*d - 3*A*b*e + a*B*e)*(a + b*x)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(d + e*
x))/(b*d - a*e)])/(b*(b*d - a*e)^2*(a + b*x)*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.51, size = 178, normalized size = 1.27 \begin {gather*} \frac {2 a A e^2-a B e (d+e x)-2 a B d e+3 A b e (d+e x)-2 A b d e+2 b B d^2-2 b B d (d+e x)}{\sqrt {d+e x} (b d-a e)^2 (-a e-b (d+e x)+b d)}+\frac {(-a B e+3 A b e-2 b B d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{\sqrt {b} (b d-a e)^2 \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 775, normalized size = 5.54 \begin {gather*} \left [-\frac {{\left (2 \, B a b d^{2} + {\left (B a^{2} - 3 \, A a b\right )} d e + {\left (2 \, B b^{2} d e + {\left (B a b - 3 \, A b^{2}\right )} e^{2}\right )} x^{2} + {\left (2 \, B b^{2} d^{2} + 3 \, {\left (B a b - A b^{2}\right )} d e + {\left (B a^{2} - 3 \, A a b\right )} e^{2}\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, A a^{2} b e^{2} + {\left (3 \, B a b^{2} - A b^{3}\right )} d^{2} - {\left (3 \, B a^{2} b + A a b^{2}\right )} d e + {\left (2 \, B b^{3} d^{2} - {\left (B a b^{2} + 3 \, A b^{3}\right )} d e - {\left (B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (a b^{4} d^{4} - 3 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} - a^{4} b d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{2} + {\left (b^{5} d^{4} - 2 \, a b^{4} d^{3} e + 2 \, a^{3} b^{2} d e^{3} - a^{4} b e^{4}\right )} x\right )}}, \frac {{\left (2 \, B a b d^{2} + {\left (B a^{2} - 3 \, A a b\right )} d e + {\left (2 \, B b^{2} d e + {\left (B a b - 3 \, A b^{2}\right )} e^{2}\right )} x^{2} + {\left (2 \, B b^{2} d^{2} + 3 \, {\left (B a b - A b^{2}\right )} d e + {\left (B a^{2} - 3 \, A a b\right )} e^{2}\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (2 \, A a^{2} b e^{2} + {\left (3 \, B a b^{2} - A b^{3}\right )} d^{2} - {\left (3 \, B a^{2} b + A a b^{2}\right )} d e + {\left (2 \, B b^{3} d^{2} - {\left (B a b^{2} + 3 \, A b^{3}\right )} d e - {\left (B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{a b^{4} d^{4} - 3 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} - a^{4} b d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{2} + {\left (b^{5} d^{4} - 2 \, a b^{4} d^{3} e + 2 \, a^{3} b^{2} d e^{3} - a^{4} b e^{4}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((2*B*a*b*d^2 + (B*a^2 - 3*A*a*b)*d*e + (2*B*b^2*d*e + (B*a*b - 3*A*b^2)*e^2)*x^2 + (2*B*b^2*d^2 + 3*(B*
a*b - A*b^2)*d*e + (B*a^2 - 3*A*a*b)*e^2)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e + 2*sqrt(b^2*d - a*b
*e)*sqrt(e*x + d))/(b*x + a)) - 2*(2*A*a^2*b*e^2 + (3*B*a*b^2 - A*b^3)*d^2 - (3*B*a^2*b + A*a*b^2)*d*e + (2*B*
b^3*d^2 - (B*a*b^2 + 3*A*b^3)*d*e - (B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(a*b^4*d^4 - 3*a^2*b^3*d^3*e
+ 3*a^3*b^2*d^2*e^2 - a^4*b*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^3*b^2*e^4)*x^2 + (b^5*d
^4 - 2*a*b^4*d^3*e + 2*a^3*b^2*d*e^3 - a^4*b*e^4)*x), ((2*B*a*b*d^2 + (B*a^2 - 3*A*a*b)*d*e + (2*B*b^2*d*e + (
B*a*b - 3*A*b^2)*e^2)*x^2 + (2*B*b^2*d^2 + 3*(B*a*b - A*b^2)*d*e + (B*a^2 - 3*A*a*b)*e^2)*x)*sqrt(-b^2*d + a*b
*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (2*A*a^2*b*e^2 + (3*B*a*b^2 - A*b^3)*d^2 - (3*B
*a^2*b + A*a*b^2)*d*e + (2*B*b^3*d^2 - (B*a*b^2 + 3*A*b^3)*d*e - (B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/
(a*b^4*d^4 - 3*a^2*b^3*d^3*e + 3*a^3*b^2*d^2*e^2 - a^4*b*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^
3 - a^3*b^2*e^4)*x^2 + (b^5*d^4 - 2*a*b^4*d^3*e + 2*a^3*b^2*d*e^3 - a^4*b*e^4)*x)]

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giac [A]  time = 0.19, size = 204, normalized size = 1.46 \begin {gather*} \frac {{\left (2 \, B b d + B a e - 3 \, A b e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, {\left (x e + d\right )} B b d - 2 \, B b d^{2} + {\left (x e + d\right )} B a e - 3 \, {\left (x e + d\right )} A b e + 2 \, B a d e + 2 \, A b d e - 2 \, A a e^{2}}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left ({\left (x e + d\right )}^{\frac {3}{2}} b - \sqrt {x e + d} b d + \sqrt {x e + d} a e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 253, normalized size = 1.81 \begin {gather*} -\frac {3 A b e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}}+\frac {B a e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}}+\frac {2 B b d \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{2} \sqrt {\left (a e -b d \right ) b}}-\frac {\sqrt {e x +d}\, A b e}{\left (a e -b d \right )^{2} \left (b e x +a e \right )}+\frac {\sqrt {e x +d}\, B a e}{\left (a e -b d \right )^{2} \left (b e x +a e \right )}-\frac {2 A e}{\left (a e -b d \right )^{2} \sqrt {e x +d}}+\frac {2 B d}{\left (a e -b d \right )^{2} \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2)/(b^2*x^2+2*a*b*x+a^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(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 = 2.09, size = 156, normalized size = 1.11 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{{\left (a\,e-b\,d\right )}^{5/2}}\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{\sqrt {b}\,{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,e-B\,d\right )}{a\,e-b\,d}-\frac {\left (d+e\,x\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^2}}{b\,{\left (d+e\,x\right )}^{3/2}+\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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