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

Optimal. Leaf size=95 \[ -\frac{\sec (c+d x) (b-a \tan (c+d x))}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{2 d \left (a^2+b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0916806, antiderivative size = 118, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3512, 721, 725, 206} \[ -\frac{\sec (c+d x) (b-a \tan (c+d x))}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{3/2} \sqrt{\sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + b*Tan[c + d*x])^3,x]

[Out]

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

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]

Rule 721

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(-2*a*e + (2*c*
d)*x)*(a + c*x^2)^p)/(2*(m + 1)*(c*d^2 + a*e^2)), x] - Dist[(4*a*c*p)/(2*(m + 1)*(c*d^2 + a*e^2)), Int[(d + e*
x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2,
0] && GtQ[p, 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 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])

Rubi steps

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

Mathematica [C]  time = 0.293105, size = 132, normalized size = 1.39 \[ \frac{\left (a^2+b^2\right ) (a \sin (c+d x)-b \cos (c+d x))+2 \sqrt{a^2+b^2} (a \cos (c+d x)+b \sin (c+d x))^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{2 d (a-i b)^2 (a+i b)^2 (a \cos (c+d x)+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.118, size = 191, normalized size = 2. \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{2}} \left ( -1/2\,{\frac{ \left ({a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{2}+{b}^{2} \right ) a}}-1/2\,{\frac{b \left ({a}^{2}-2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{ \left ({a}^{2}+{b}^{2} \right ){a}^{2}}}-1/2\,{\frac{ \left ({a}^{2}-2\,{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ({a}^{2}+{b}^{2} \right ) a}}+1/2\,{\frac{b}{{a}^{2}+{b}^{2}}} \right ) }+{{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ) \left ({a}^{2}+{b}^{2} \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2*(-1/2*(a^2+2*b^2)/(a^2+b^2)/a*tan(1/2*d*x+1/2*c)^3-1/2*b*(a^2-2*b^2)/(a^2+b^2)/a^2*tan(1/2*d*x+1/2*c)^
2-1/2*(a^2-2*b^2)/(a^2+b^2)/a*tan(1/2*d*x+1/2*c)+1/2*b/(a^2+b^2))/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)
*b-a)^2+1/(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.01829, size = 679, normalized size = 7.15 \begin{align*} \frac{{\left (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 )} \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 (a^{2} b + b^{3}\right )} \cos \left (d x + c\right ) + 2 \,{\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**3/(a + b*tan(c + d*x))**3, x)

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Giac [B]  time = 1.91071, size = 298, normalized size = 3.14 \begin{align*} -\frac{\frac{\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 (a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2} b\right )}}{{\left (a^{4} + a^{2} 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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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