3.21 \(\int \frac {x}{(a+b \tan (c+d x^2))^2} \, dx\)

Optimal. Leaf size=94 \[ -\frac {b}{2 d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}+\frac {a b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{d \left (a^2+b^2\right )^2}+\frac {x^2 \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2} \]

[Out]

1/2*(a^2-b^2)*x^2/(a^2+b^2)^2+a*b*ln(a*cos(d*x^2+c)+b*sin(d*x^2+c))/(a^2+b^2)^2/d-1/2*b/(a^2+b^2)/d/(a+b*tan(d
*x^2+c))

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Rubi [A]  time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3747, 3483, 3531, 3530} \[ -\frac {b}{2 d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d x^2\right )\right )}+\frac {a b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{d \left (a^2+b^2\right )^2}+\frac {x^2 \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Tan[c + d*x^2])^2,x]

[Out]

((a^2 - b^2)*x^2)/(2*(a^2 + b^2)^2) + (a*b*Log[a*Cos[c + d*x^2] + b*Sin[c + d*x^2]])/((a^2 + b^2)^2*d) - b/(2*
(a^2 + b^2)*d*(a + b*Tan[c + d*x^2]))

Rule 3483

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

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

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \tan \left (c+d x^2\right )\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b \tan (c+d x))^2} \, dx,x,x^2\right )\\ &=-\frac {b}{2 \left (a^2+b^2\right ) d \left (a+b \tan \left (c+d x^2\right )\right )}+\frac {\operatorname {Subst}\left (\int \frac {a-b \tan (c+d x)}{a+b \tan (c+d x)} \, dx,x,x^2\right )}{2 \left (a^2+b^2\right )}\\ &=\frac {\left (a^2-b^2\right ) x^2}{2 \left (a^2+b^2\right )^2}-\frac {b}{2 \left (a^2+b^2\right ) d \left (a+b \tan \left (c+d x^2\right )\right )}+\frac {(a b) \operatorname {Subst}\left (\int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx,x,x^2\right )}{\left (a^2+b^2\right )^2}\\ &=\frac {\left (a^2-b^2\right ) x^2}{2 \left (a^2+b^2\right )^2}+\frac {a b \log \left (a \cos \left (c+d x^2\right )+b \sin \left (c+d x^2\right )\right )}{\left (a^2+b^2\right )^2 d}-\frac {b}{2 \left (a^2+b^2\right ) d \left (a+b \tan \left (c+d x^2\right )\right )}\\ \end {align*}

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Mathematica [C]  time = 1.30, size = 114, normalized size = 1.21 \[ \frac {\frac {2 b \left (2 a \log \left (a+b \tan \left (c+d x^2\right )\right )-\frac {a^2+b^2}{a+b \tan \left (c+d x^2\right )}\right )}{\left (a^2+b^2\right )^2}-\frac {i \log \left (-\tan \left (c+d x^2\right )+i\right )}{(a+i b)^2}+\frac {i \log \left (\tan \left (c+d x^2\right )+i\right )}{(a-i b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Tan[c + d*x^2])^2,x]

[Out]

(((-I)*Log[I - Tan[c + d*x^2]])/(a + I*b)^2 + (I*Log[I + Tan[c + d*x^2]])/(a - I*b)^2 + (2*b*(2*a*Log[a + b*Ta
n[c + d*x^2]] - (a^2 + b^2)/(a + b*Tan[c + d*x^2])))/(a^2 + b^2)^2)/(4*d)

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fricas [A]  time = 0.52, size = 169, normalized size = 1.80 \[ \frac {{\left (a^{3} - a b^{2}\right )} d x^{2} - b^{3} + {\left (a b^{2} \tan \left (d x^{2} + c\right ) + a^{2} b\right )} \log \left (\frac {b^{2} \tan \left (d x^{2} + c\right )^{2} + 2 \, a b \tan \left (d x^{2} + c\right ) + a^{2}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) + {\left ({\left (a^{2} b - b^{3}\right )} d x^{2} + a b^{2}\right )} \tan \left (d x^{2} + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x^{2} + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^3 - a*b^2)*d*x^2 - b^3 + (a*b^2*tan(d*x^2 + c) + a^2*b)*log((b^2*tan(d*x^2 + c)^2 + 2*a*b*tan(d*x^2 +
c) + a^2)/(tan(d*x^2 + c)^2 + 1)) + ((a^2*b - b^3)*d*x^2 + a*b^2)*tan(d*x^2 + c))/((a^4*b + 2*a^2*b^3 + b^5)*d
*tan(d*x^2 + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d)

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giac [A]  time = 1.42, size = 159, normalized size = 1.69 \[ \frac {a b^{2} \log \left ({\left | b \tan \left (d x^{2} + c\right ) + a \right |}\right )}{a^{4} b d + 2 \, a^{2} b^{3} d + b^{5} d} - \frac {a b \log \left (\tan \left (d x^{2} + c\right )^{2} + 1\right )}{2 \, {\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} + \frac {{\left (d x^{2} + c\right )} {\left (a^{2} - b^{2}\right )}}{2 \, {\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac {a^{2} b + b^{3}}{2 \, {\left (a^{2} + b^{2}\right )}^{2} {\left (b \tan \left (d x^{2} + c\right ) + a\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b^2*log(abs(b*tan(d*x^2 + c) + a))/(a^4*b*d + 2*a^2*b^3*d + b^5*d) - 1/2*a*b*log(tan(d*x^2 + c)^2 + 1)/(a^4*
d + 2*a^2*b^2*d + b^4*d) + 1/2*(d*x^2 + c)*(a^2 - b^2)/(a^4*d + 2*a^2*b^2*d + b^4*d) - 1/2*(a^2*b + b^3)/((a^2
+ b^2)^2*(b*tan(d*x^2 + c) + a)*d)

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maple [A]  time = 0.16, size = 140, normalized size = 1.49 \[ -\frac {b}{2 \left (a^{2}+b^{2}\right ) d \left (a +b \tan \left (d \,x^{2}+c \right )\right )}+\frac {a b \ln \left (a +b \tan \left (d \,x^{2}+c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {a b \ln \left (1+\tan ^{2}\left (d \,x^{2}+c \right )\right )}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \left (d \,x^{2}+c \right )\right ) a^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (d \,x^{2}+c \right )\right ) b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/(a^2+b^2)/d/(a+b*tan(d*x^2+c))+1/d*a*b/(a^2+b^2)^2*ln(a+b*tan(d*x^2+c))-1/2/d/(a^2+b^2)^2*a*b*ln(1+tan(
d*x^2+c)^2)+1/2/d/(a^2+b^2)^2*arctan(tan(d*x^2+c))*a^2-1/2/d/(a^2+b^2)^2*arctan(tan(d*x^2+c))*b^2

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maxima [B]  time = 1.41, size = 556, normalized size = 5.91 \[ \frac {{\left (a^{4} - b^{4}\right )} d x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{4} - b^{4}\right )} d x^{2} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{4} - b^{4}\right )} d x^{2} - 2 \, {\left (2 \, a b^{3} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right ) + {\left (4 \, a^{2} b^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) + a^{3} b + a b^{3} + {\left (a^{3} b + a b^{3}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{3} b + a b^{3}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, c\right )^{2} + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, c\right )^{2}}\right ) + 2 \, {\left (a^{2} b^{2} - b^{4} + 2 \, {\left (a^{3} b - a b^{3}\right )} d x^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{2 \, {\left ({\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, {\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} d \cos \left (2 \, d x^{2} + 2 \, c\right ) + 4 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (2 \, d x^{2} + 2 \, c\right ) + {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^4 - b^4)*d*x^2*cos(2*d*x^2 + 2*c)^2 + (a^4 - b^4)*d*x^2*sin(2*d*x^2 + 2*c)^2 + (a^4 - b^4)*d*x^2 - 2*(
2*a*b^3 - (a^4 - 2*a^2*b^2 + b^4)*d*x^2)*cos(2*d*x^2 + 2*c) + (4*a^2*b^2*sin(2*d*x^2 + 2*c) + a^3*b + a*b^3 +
(a^3*b + a*b^3)*cos(2*d*x^2 + 2*c)^2 + (a^3*b + a*b^3)*sin(2*d*x^2 + 2*c)^2 + 2*(a^3*b - a*b^3)*cos(2*d*x^2 +
2*c))*log(((a^2 + b^2)*cos(2*d*x^2 + 2*c)^2 + 4*a*b*sin(2*d*x^2 + 2*c) + (a^2 + b^2)*sin(2*d*x^2 + 2*c)^2 + a^
2 + b^2 + 2*(a^2 - b^2)*cos(2*d*x^2 + 2*c))/((a^2 + b^2)*cos(2*c)^2 + (a^2 + b^2)*sin(2*c)^2)) + 2*(a^2*b^2 -
b^4 + 2*(a^3*b - a*b^3)*d*x^2)*sin(2*d*x^2 + 2*c))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d*cos(2*d*x^2 + 2*c)^2
+ (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d*sin(2*d*x^2 + 2*c)^2 + 2*(a^6 + a^4*b^2 - a^2*b^4 - b^6)*d*cos(2*d*x^
2 + 2*c) + 4*(a^5*b + 2*a^3*b^3 + a*b^5)*d*sin(2*d*x^2 + 2*c) + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)

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mupad [B]  time = 3.60, size = 173, normalized size = 1.84 \[ \frac {\frac {x^2\,\mathrm {tan}\left (d\,x^2+c\right )\,\left (\frac {a^2\,b}{2}-\frac {b^3}{2}\right )}{{\left (a^2+b^2\right )}^2}-\frac {x^2\,\left (\frac {a\,b^2}{2}-\frac {a^3}{2}\right )}{{\left (a^2+b^2\right )}^2}+\frac {b^2\,\mathrm {tan}\left (d\,x^2+c\right )}{2\,a\,d\,\left (a^2+b^2\right )}}{a+b\,\mathrm {tan}\left (d\,x^2+c\right )}-\frac {a\,b\,\ln \left ({\mathrm {tan}\left (d\,x^2+c\right )}^2+1\right )}{2\,\left (d\,a^4+2\,d\,a^2\,b^2+d\,b^4\right )}+\frac {a\,b\,\ln \left (a+b\,\mathrm {tan}\left (d\,x^2+c\right )\right )}{d\,{\left (a^2+b^2\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*tan(c + d*x^2))^2,x)

[Out]

((x^2*tan(c + d*x^2)*((a^2*b)/2 - b^3/2))/(a^2 + b^2)^2 - (x^2*((a*b^2)/2 - a^3/2))/(a^2 + b^2)^2 + (b^2*tan(c
+ d*x^2))/(2*a*d*(a^2 + b^2)))/(a + b*tan(c + d*x^2)) - (a*b*log(tan(c + d*x^2)^2 + 1))/(2*(a^4*d + b^4*d + 2
*a^2*b^2*d)) + (a*b*log(a + b*tan(c + d*x^2)))/(d*(a^2 + b^2)^2)

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sympy [A]  time = 1.56, size = 1584, normalized size = 16.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**2/tan(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-(atan(tan(c + d*x**2)) + pi*floor((c + d*x**
2 - pi/2)/pi))*tan(c + d*x**2)**2/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + 2*I
*(atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2)**2 - 16*
I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + (atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))/(8*b**2*d*ta
n(c + d*x**2)**2 - 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) - tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2)**2 - 16
*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + 2*I/(8*b**2*d*tan(c + d*x**2)**2 - 16*I*b**2*d*tan(c + d*x**2) - 8*b**
2*d), Eq(a, -I*b)), (-(atan(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))*tan(c + d*x**2)**2/(8*b**2*d*
tan(c + d*x**2)**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) - 2*I*(atan(tan(c + d*x**2)) + pi*floor((c + d*x*
*2 - pi/2)/pi))*tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2)**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) + (atan
(tan(c + d*x**2)) + pi*floor((c + d*x**2 - pi/2)/pi))/(8*b**2*d*tan(c + d*x**2)**2 + 16*I*b**2*d*tan(c + d*x**
2) - 8*b**2*d) - tan(c + d*x**2)/(8*b**2*d*tan(c + d*x**2)**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d) - 2*I/
(8*b**2*d*tan(c + d*x**2)**2 + 16*I*b**2*d*tan(c + d*x**2) - 8*b**2*d), Eq(a, I*b)), (x**2/(2*(a + b*tan(c))**
2), Eq(d, 0)), (x**2/(2*a**2), Eq(b, 0)), (a**3*d*x**2/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d
+ 4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)) + a**2*b*d*x**2*tan(c + d*x**2)/(2*a*
*5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan(
c + d*x**2)) + 2*a**2*b*log(a/b + tan(c + d*x**2))/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*
a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)) - a**2*b*log(tan(c + d*x**2)**2 + 1)/(2*a
**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan
(c + d*x**2)) - a**2*b/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x**2)
+ 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)) - a*b**2*d*x**2/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*
d + 4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)) + 2*a*b**2*log(a/b + tan(c + d*x**2
))*tan(c + d*x**2)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x**2) + 2*
a*b**4*d + 2*b**5*d*tan(c + d*x**2)) - a*b**2*log(tan(c + d*x**2)**2 + 1)*tan(c + d*x**2)/(2*a**5*d + 2*a**4*b
*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)) -
b**3*d*x**2*tan(c + d*x**2)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x
**2) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)) - b**3/(2*a**5*d + 2*a**4*b*d*tan(c + d*x**2) + 4*a**3*b**2*d +
4*a**2*b**3*d*tan(c + d*x**2) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x**2)), True))

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