3.87 \(\int \frac {1}{(a+b (c+d x)^2)^3} \, dx\)

Optimal. Leaf size=91 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} d}+\frac {3 (c+d x)}{8 a^2 d \left (a+b (c+d x)^2\right )}+\frac {c+d x}{4 a d \left (a+b (c+d x)^2\right )^2} \]

[Out]

1/4*(d*x+c)/a/d/(a+b*(d*x+c)^2)^2+3/8*(d*x+c)/a^2/d/(a+b*(d*x+c)^2)+3/8*arctan((d*x+c)*b^(1/2)/a^(1/2))/a^(5/2
)/d/b^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {247, 199, 205} \[ \frac {3 (c+d x)}{8 a^2 d \left (a+b (c+d x)^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} d}+\frac {c+d x}{4 a d \left (a+b (c+d x)^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c + d*x)^2)^(-3),x]

[Out]

(c + d*x)/(4*a*d*(a + b*(c + d*x)^2)^2) + (3*(c + d*x))/(8*a^2*d*(a + b*(c + d*x)^2)) + (3*ArcTan[(Sqrt[b]*(c
+ d*x))/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]*d)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b (c+d x)^2\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {c+d x}{4 a d \left (a+b (c+d x)^2\right )^2}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^2} \, dx,x,c+d x\right )}{4 a d}\\ &=\frac {c+d x}{4 a d \left (a+b (c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 a^2 d \left (a+b (c+d x)^2\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,c+d x\right )}{8 a^2 d}\\ &=\frac {c+d x}{4 a d \left (a+b (c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 a^2 d \left (a+b (c+d x)^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 75, normalized size = 0.82 \[ \frac {\frac {\sqrt {a} (c+d x) \left (5 a+3 b (c+d x)^2\right )}{\left (a+b (c+d x)^2\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} (c+d x)}{\sqrt {a}}\right )}{\sqrt {b}}}{8 a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c + d*x)^2)^(-3),x]

[Out]

((Sqrt[a]*(c + d*x)*(5*a + 3*b*(c + d*x)^2))/(a + b*(c + d*x)^2)^2 + (3*ArcTan[(Sqrt[b]*(c + d*x))/Sqrt[a]])/S
qrt[b])/(8*a^(5/2)*d)

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fricas [B]  time = 1.01, size = 595, normalized size = 6.54 \[ \left [\frac {6 \, a b^{2} d^{3} x^{3} + 18 \, a b^{2} c d^{2} x^{2} + 6 \, a b^{2} c^{3} + 10 \, a^{2} b c + 2 \, {\left (9 \, a b^{2} c^{2} + 5 \, a^{2} b\right )} d x - 3 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + b^{2} c^{4} + 2 \, {\left (3 \, b^{2} c^{2} + a b\right )} d^{2} x^{2} + 2 \, a b c^{2} + 4 \, {\left (b^{2} c^{3} + a b c\right )} d x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b d^{2} x^{2} + 2 \, b c d x + b c^{2} - 2 \, \sqrt {-a b} {\left (d x + c\right )} - a}{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}\right )}{16 \, {\left (a^{3} b^{3} d^{5} x^{4} + 4 \, a^{3} b^{3} c d^{4} x^{3} + 2 \, {\left (3 \, a^{3} b^{3} c^{2} + a^{4} b^{2}\right )} d^{3} x^{2} + 4 \, {\left (a^{3} b^{3} c^{3} + a^{4} b^{2} c\right )} d^{2} x + {\left (a^{3} b^{3} c^{4} + 2 \, a^{4} b^{2} c^{2} + a^{5} b\right )} d\right )}}, \frac {3 \, a b^{2} d^{3} x^{3} + 9 \, a b^{2} c d^{2} x^{2} + 3 \, a b^{2} c^{3} + 5 \, a^{2} b c + {\left (9 \, a b^{2} c^{2} + 5 \, a^{2} b\right )} d x + 3 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + b^{2} c^{4} + 2 \, {\left (3 \, b^{2} c^{2} + a b\right )} d^{2} x^{2} + 2 \, a b c^{2} + 4 \, {\left (b^{2} c^{3} + a b c\right )} d x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} {\left (d x + c\right )}}{a}\right )}{8 \, {\left (a^{3} b^{3} d^{5} x^{4} + 4 \, a^{3} b^{3} c d^{4} x^{3} + 2 \, {\left (3 \, a^{3} b^{3} c^{2} + a^{4} b^{2}\right )} d^{3} x^{2} + 4 \, {\left (a^{3} b^{3} c^{3} + a^{4} b^{2} c\right )} d^{2} x + {\left (a^{3} b^{3} c^{4} + 2 \, a^{4} b^{2} c^{2} + a^{5} b\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

[1/16*(6*a*b^2*d^3*x^3 + 18*a*b^2*c*d^2*x^2 + 6*a*b^2*c^3 + 10*a^2*b*c + 2*(9*a*b^2*c^2 + 5*a^2*b)*d*x - 3*(b^
2*d^4*x^4 + 4*b^2*c*d^3*x^3 + b^2*c^4 + 2*(3*b^2*c^2 + a*b)*d^2*x^2 + 2*a*b*c^2 + 4*(b^2*c^3 + a*b*c)*d*x + a^
2)*sqrt(-a*b)*log((b*d^2*x^2 + 2*b*c*d*x + b*c^2 - 2*sqrt(-a*b)*(d*x + c) - a)/(b*d^2*x^2 + 2*b*c*d*x + b*c^2
+ a)))/(a^3*b^3*d^5*x^4 + 4*a^3*b^3*c*d^4*x^3 + 2*(3*a^3*b^3*c^2 + a^4*b^2)*d^3*x^2 + 4*(a^3*b^3*c^3 + a^4*b^2
*c)*d^2*x + (a^3*b^3*c^4 + 2*a^4*b^2*c^2 + a^5*b)*d), 1/8*(3*a*b^2*d^3*x^3 + 9*a*b^2*c*d^2*x^2 + 3*a*b^2*c^3 +
5*a^2*b*c + (9*a*b^2*c^2 + 5*a^2*b)*d*x + 3*(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + b^2*c^4 + 2*(3*b^2*c^2 + a*b)*d^
2*x^2 + 2*a*b*c^2 + 4*(b^2*c^3 + a*b*c)*d*x + a^2)*sqrt(a*b)*arctan(sqrt(a*b)*(d*x + c)/a))/(a^3*b^3*d^5*x^4 +
4*a^3*b^3*c*d^4*x^3 + 2*(3*a^3*b^3*c^2 + a^4*b^2)*d^3*x^2 + 4*(a^3*b^3*c^3 + a^4*b^2*c)*d^2*x + (a^3*b^3*c^4
+ 2*a^4*b^2*c^2 + a^5*b)*d)]

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giac [A]  time = 0.41, size = 103, normalized size = 1.13 \[ \frac {3 \, \arctan \left (\frac {b d x + b c}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} d} + \frac {3 \, b d^{3} x^{3} + 9 \, b c d^{2} x^{2} + 9 \, b c^{2} d x + 3 \, b c^{3} + 5 \, a d x + 5 \, a c}{8 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}^{2} a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)^2)^3,x, algorithm="giac")

[Out]

3/8*arctan((b*d*x + b*c)/sqrt(a*b))/(sqrt(a*b)*a^2*d) + 1/8*(3*b*d^3*x^3 + 9*b*c*d^2*x^2 + 9*b*c^2*d*x + 3*b*c
^3 + 5*a*d*x + 5*a*c)/((b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a)^2*a^2*d)

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maple [A]  time = 0.00, size = 147, normalized size = 1.62 \[ \frac {3 x}{8 \left (b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a \right ) a^{2}}+\frac {3 c}{8 \left (b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a \right ) a^{2} d}+\frac {3 \arctan \left (\frac {2 b \,d^{2} x +2 b d c}{2 \sqrt {a b}\, d}\right )}{8 \sqrt {a b}\, a^{2} d}+\frac {2 b \,d^{2} x +2 b d c}{8 \left (b \,d^{2} x^{2}+2 b c d x +b \,c^{2}+a \right )^{2} a b \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*(d*x+c)^2)^3,x)

[Out]

1/8*(2*b*d^2*x+2*b*c*d)/a/b/d^2/(b*d^2*x^2+2*b*c*d*x+b*c^2+a)^2+3/8/a^2/(b*d^2*x^2+2*b*c*d*x+b*c^2+a)*x+3/8/a^
2/d/(b*d^2*x^2+2*b*c*d*x+b*c^2+a)*c+3/8/a^2/d/(a*b)^(1/2)*arctan(1/2*(2*b*d^2*x+2*b*c*d)/(a*b)^(1/2)/d)

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maxima [B]  time = 1.59, size = 184, normalized size = 2.02 \[ \frac {3 \, b d^{3} x^{3} + 9 \, b c d^{2} x^{2} + 3 \, b c^{3} + {\left (9 \, b c^{2} + 5 \, a\right )} d x + 5 \, a c}{8 \, {\left (a^{2} b^{2} d^{5} x^{4} + 4 \, a^{2} b^{2} c d^{4} x^{3} + 2 \, {\left (3 \, a^{2} b^{2} c^{2} + a^{3} b\right )} d^{3} x^{2} + 4 \, {\left (a^{2} b^{2} c^{3} + a^{3} b c\right )} d^{2} x + {\left (a^{2} b^{2} c^{4} + 2 \, a^{3} b c^{2} + a^{4}\right )} d\right )}} + \frac {3 \, \arctan \left (\frac {b d^{2} x + b c d}{\sqrt {a b} d}\right )}{8 \, \sqrt {a b} a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/8*(3*b*d^3*x^3 + 9*b*c*d^2*x^2 + 3*b*c^3 + (9*b*c^2 + 5*a)*d*x + 5*a*c)/(a^2*b^2*d^5*x^4 + 4*a^2*b^2*c*d^4*x
^3 + 2*(3*a^2*b^2*c^2 + a^3*b)*d^3*x^2 + 4*(a^2*b^2*c^3 + a^3*b*c)*d^2*x + (a^2*b^2*c^4 + 2*a^3*b*c^2 + a^4)*d
) + 3/8*arctan((b*d^2*x + b*c*d)/(sqrt(a*b)*d))/(sqrt(a*b)*a^2*d)

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mupad [B]  time = 2.22, size = 181, normalized size = 1.99 \[ \frac {\frac {x\,\left (9\,b\,c^2+5\,a\right )}{8\,a^2}+\frac {3\,b\,c^3+5\,a\,c}{8\,a^2\,d}+\frac {3\,b\,d^2\,x^3}{8\,a^2}+\frac {9\,b\,c\,d\,x^2}{8\,a^2}}{x^2\,\left (6\,b^2\,c^2\,d^2+2\,a\,b\,d^2\right )+x\,\left (4\,d\,b^2\,c^3+4\,a\,d\,b\,c\right )+a^2+b^2\,c^4+b^2\,d^4\,x^4+2\,a\,b\,c^2+4\,b^2\,c\,d^3\,x^3}+\frac {3\,\mathrm {atan}\left (\frac {8\,a^2\,\left (\frac {3\,\sqrt {b}\,c}{8\,a^{5/2}}+\frac {3\,\sqrt {b}\,d\,x}{8\,a^{5/2}}\right )}{3}\right )}{8\,a^{5/2}\,\sqrt {b}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*(c + d*x)^2)^3,x)

[Out]

((x*(5*a + 9*b*c^2))/(8*a^2) + (5*a*c + 3*b*c^3)/(8*a^2*d) + (3*b*d^2*x^3)/(8*a^2) + (9*b*c*d*x^2)/(8*a^2))/(x
^2*(6*b^2*c^2*d^2 + 2*a*b*d^2) + x*(4*b^2*c^3*d + 4*a*b*c*d) + a^2 + b^2*c^4 + b^2*d^4*x^4 + 2*a*b*c^2 + 4*b^2
*c*d^3*x^3) + (3*atan((8*a^2*((3*b^(1/2)*c)/(8*a^(5/2)) + (3*b^(1/2)*d*x)/(8*a^(5/2))))/3))/(8*a^(5/2)*b^(1/2)
*d)

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sympy [B]  time = 1.25, size = 257, normalized size = 2.82 \[ \frac {5 a c + 3 b c^{3} + 9 b c d^{2} x^{2} + 3 b d^{3} x^{3} + x \left (5 a d + 9 b c^{2} d\right )}{8 a^{4} d + 16 a^{3} b c^{2} d + 8 a^{2} b^{2} c^{4} d + 32 a^{2} b^{2} c d^{4} x^{3} + 8 a^{2} b^{2} d^{5} x^{4} + x^{2} \left (16 a^{3} b d^{3} + 48 a^{2} b^{2} c^{2} d^{3}\right ) + x \left (32 a^{3} b c d^{2} + 32 a^{2} b^{2} c^{3} d^{2}\right )} + \frac {- \frac {3 \sqrt {- \frac {1}{a^{5} b}} \log {\left (x + \frac {- 3 a^{3} \sqrt {- \frac {1}{a^{5} b}} + 3 c}{3 d} \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{5} b}} \log {\left (x + \frac {3 a^{3} \sqrt {- \frac {1}{a^{5} b}} + 3 c}{3 d} \right )}}{16}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)**2)**3,x)

[Out]

(5*a*c + 3*b*c**3 + 9*b*c*d**2*x**2 + 3*b*d**3*x**3 + x*(5*a*d + 9*b*c**2*d))/(8*a**4*d + 16*a**3*b*c**2*d + 8
*a**2*b**2*c**4*d + 32*a**2*b**2*c*d**4*x**3 + 8*a**2*b**2*d**5*x**4 + x**2*(16*a**3*b*d**3 + 48*a**2*b**2*c**
2*d**3) + x*(32*a**3*b*c*d**2 + 32*a**2*b**2*c**3*d**2)) + (-3*sqrt(-1/(a**5*b))*log(x + (-3*a**3*sqrt(-1/(a**
5*b)) + 3*c)/(3*d))/16 + 3*sqrt(-1/(a**5*b))*log(x + (3*a**3*sqrt(-1/(a**5*b)) + 3*c)/(3*d))/16)/d

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