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

Optimal. Leaf size=82 \[ -\frac {b \sec (c+d x)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {a \tanh ^{-1}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}} \]

[Out]

-a*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)/d-b*sec(d*x+c)/(a^2+b^2)/d/(a+b*tan(d*
x+c))

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Rubi [A]  time = 0.07, antiderivative size = 105, normalized size of antiderivative = 1.28, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3512, 731, 725, 206} \[ -\frac {b \sec (c+d x)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {a \sec (c+d x) \tanh ^{-1}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{3/2} \sqrt {\sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Sec[c + d*x])/((a^2 + b^2)^(3/2)*d*S
qrt[Sec[c + d*x]^2])) - (b*Sec[c + d*x])/((a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 78, normalized size = 0.95 \[ \frac {\frac {2 a \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {b \sec (c+d x)}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2) - (b*Sec[c + d*x])/((a^2 + b^2)*(a
+ b*Tan[c + d*x])))/d

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fricas [B]  time = 0.67, size = 215, normalized size = 2.62 \[ -\frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a^{2} \cos \left (d x + c\right ) + a b \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.94, size = 138, normalized size = 1.68 \[ -\frac {\frac {a \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b\right )}}{{\left (a^{3} + a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 118, normalized size = 1.44 \[ \frac {-\frac {2 \left (-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}+b^{2}\right ) a}-\frac {b}{a^{2}+b^{2}}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a}+\frac {2 a \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.69, size = 182, normalized size = 2.22 \[ -\frac {\frac {a \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a b + \frac {b^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}}{a^{4} + a^{2} b^{2} + \frac {2 \, {\left (a^{3} b + a b^{3}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (a^{4} + a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqr
t(a^2 + b^2)))/(a^2 + b^2)^(3/2) + 2*(a*b + b^2*sin(d*x + c)/(cos(d*x + c) + 1))/(a^4 + a^2*b^2 + 2*(a^3*b + a
*b^3)*sin(d*x + c)/(cos(d*x + c) + 1) - (a^4 + a^2*b^2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2))/d

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mupad [B]  time = 3.98, size = 136, normalized size = 1.66 \[ -\frac {\frac {2\,b}{a^2+b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,\left (a^2+b^2\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {a\,\mathrm {atan}\left (\frac {a^2\,b\,1{}\mathrm {i}+b^3\,1{}\mathrm {i}-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+b^2\right )\,1{}\mathrm {i}}{{\left (a^2+b^2\right )}^{3/2}}\right )\,2{}\mathrm {i}}{d\,{\left (a^2+b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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