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

Optimal. Leaf size=122 \[ -\frac {2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

[Out]

a*(a^2-3*b^2)*x/(a^2+b^2)^3+b*(3*a^2-b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^3/d-1/2*b/(a^2+b^2)/d/(a+b*t
an(d*x+c))^2-2*a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3483, 3529, 3531, 3530} \[ -\frac {2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-3),x]

[Out]

(a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3 + (b*(3*a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^3*d) -
b/(2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (2*a*b)/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

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 3529

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \tan (c+d x))^3} \, dx &=-\frac {b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {a-b \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx}{a^2+b^2}\\ &=-\frac {b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {a^2-b^2-2 a b \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac {b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (b \left (3 a^2-b^2\right )\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=\frac {a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 1.81, size = 127, normalized size = 1.04 \[ \frac {\frac {b \left (\left (6 a^2-2 b^2\right ) \log (a+b \tan (c+d x))-\frac {\left (a^2+b^2\right ) \left (5 a^2+4 a b \tan (c+d x)+b^2\right )}{(a+b \tan (c+d x))^2}\right )}{\left (a^2+b^2\right )^3}+\frac {\log (-\tan (c+d x)+i)}{(b-i a)^3}+\frac {\log (\tan (c+d x)+i)}{(b+i a)^3}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(-3),x]

[Out]

(Log[I - Tan[c + d*x]]/((-I)*a + b)^3 + Log[I + Tan[c + d*x]]/(I*a + b)^3 + (b*((6*a^2 - 2*b^2)*Log[a + b*Tan[
c + d*x]] - ((a^2 + b^2)*(5*a^2 + b^2 + 4*a*b*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2))/(a^2 + b^2)^3)/(2*d)

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fricas [B]  time = 0.62, size = 321, normalized size = 2.63 \[ -\frac {7 \, a^{2} b^{3} + b^{5} - 2 \, {\left (a^{5} - 3 \, a^{3} b^{2}\right )} d x - {\left (5 \, a^{2} b^{3} - b^{5} + 2 \, {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} d x\right )} \tan \left (d x + c\right )^{2} - {\left (3 \, a^{4} b - a^{2} b^{3} + {\left (3 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b^{2} - a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (3 \, a^{3} b^{2} - 3 \, a b^{4} + 2 \, {\left (a^{4} b - 3 \, a^{2} b^{3}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(7*a^2*b^3 + b^5 - 2*(a^5 - 3*a^3*b^2)*d*x - (5*a^2*b^3 - b^5 + 2*(a^3*b^2 - 3*a*b^4)*d*x)*tan(d*x + c)^2
- (3*a^4*b - a^2*b^3 + (3*a^2*b^3 - b^5)*tan(d*x + c)^2 + 2*(3*a^3*b^2 - a*b^4)*tan(d*x + c))*log((b^2*tan(d*
x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) - 2*(3*a^3*b^2 - 3*a*b^4 + 2*(a^4*b - 3*a^2*b^3)*d*
x)*tan(d*x + c))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5
+ a*b^7)*d*tan(d*x + c) + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*d)

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giac [B]  time = 0.59, size = 265, normalized size = 2.17 \[ \frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (3 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac {9 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 3 \, b^{5} \tan \left (d x + c\right )^{2} + 22 \, a^{3} b^{2} \tan \left (d x + c\right ) - 2 \, a b^{4} \tan \left (d x + c\right ) + 14 \, a^{4} b + 3 \, a^{2} b^{3} + b^{5}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1)
/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(3*a^2*b^2 - b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3
*a^2*b^5 + b^7) - (9*a^2*b^3*tan(d*x + c)^2 - 3*b^5*tan(d*x + c)^2 + 22*a^3*b^2*tan(d*x + c) - 2*a*b^4*tan(d*x
+ c) + 14*a^4*b + 3*a^2*b^3 + b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(b*tan(d*x + c) + a)^2))/d

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maple [A]  time = 0.17, size = 219, normalized size = 1.80 \[ -\frac {b}{2 \left (a^{2}+b^{2}\right ) d \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {3 a^{2} b \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {b^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {2 a b}{\left (a^{2}+b^{2}\right )^{2} d \left (a +b \tan \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3}}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.61, size = 248, normalized size = 2.03 \[ \frac {\frac {2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {4 \, a b^{2} \tan \left (d x + c\right ) + 5 \, a^{2} b + b^{3}}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.31, size = 215, normalized size = 1.76 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}-\frac {\frac {5\,a^2\,b+b^3}{2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{a^4+2\,a^2\,b^2+b^4}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {b\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^2-b^2\right )}{d\,{\left (a^2+b^2\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x))^3,x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: AttributeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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