3.2135 \(\int \frac {a+b x}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=162 \[ -\frac {3 e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}+\frac {3 \sqrt {b} e (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 21, 51, 63, 208} \[ -\frac {3 e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}+\frac {3 \sqrt {b} e (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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

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Mathematica [C]  time = 0.02, size = 64, normalized size = 0.40 \[ -\frac {2 e (a+b x) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{\sqrt {(a+b x)^2} \sqrt {d+e x} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.36, size = 423, normalized size = 2.61 \[ \left [\frac {3 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, b e x + b d + 2 \, a e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, b e x + b d + 2 \, a e\right )} \sqrt {e x + d}}{a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + {\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x}\right ] \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.27, size = 289, normalized size = 1.78 \[ -\frac {3 \, b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{{\left (b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {3 \, {\left (x e + d\right )} b e^{2} - 2 \, b d e^{2} + 2 \, a e^{3}}{{\left (b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left ({\left (x e + d\right )}^{\frac {3}{2}} b - \sqrt {x e + d} b d + \sqrt {x e + d} a e\right )}} \]

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)^(3/2),x, algorithm="giac")

[Out]

-3*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^2*d^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 2*a*b*d
*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^2*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) -
(3*(x*e + d)*b*e^2 - 2*b*d*e^2 + 2*a*e^3)/((b^2*d^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 2*a*b*d*e^2*sgn((x*
e + d)*b*e - b*d*e + a*e^2) + a^2*e^3*sgn((x*e + d)*b*e - b*d*e + a*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 [A]  time = 0.07, size = 167, normalized size = 1.03 \[ -\frac {\left (3 \sqrt {e x +d}\, b^{2} e x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 \sqrt {e x +d}\, a b e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 \sqrt {\left (a e -b d \right ) b}\, b e x +2 \sqrt {\left (a e -b d \right ) b}\, a e +\sqrt {\left (a e -b d \right ) b}\, b d \right ) \left (b x +a \right )^{2}}{\sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \]

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)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,x}{{\left (d+e\,x\right )}^{3/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b x}{\left (d + e x\right )^{\frac {3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

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)**(3/2),x)

[Out]

Integral((a + b*x)/((d + e*x)**(3/2)*((a + b*x)**2)**(3/2)), x)

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