\(\int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) [457]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {(a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3756, 398, 211} \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {(a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2} d}-\frac {(a-2 b) \tan (c+d x)}{b^2 d}+\frac {\tan ^3(c+d x)}{3 b d} \]

[In]

Int[Sec[c + d*x]^6/(a + b*Tan[c + d*x]^2),x]

[Out]

((a - b)^2*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*b^(5/2)*d) - ((a - 2*b)*Tan[c + d*x])/(b^2*d) + Ta
n[c + d*x]^3/(3*b*d)

Rule 211

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

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3756

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {3 (a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a}}+\sqrt {b} \left (-3 a+5 b+b \sec ^2(c+d x)\right ) \tan (c+d x)}{3 b^{5/2} d} \]

[In]

Integrate[Sec[c + d*x]^6/(a + b*Tan[c + d*x]^2),x]

[Out]

((3*(a - b)^2*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a] + Sqrt[b]*(-3*a + 5*b + b*Sec[c + d*x]^2)*Tan[c
+ d*x])/(3*b^(5/2)*d)

Maple [A] (verified)

Time = 12.51 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )-2 b \tan \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}}{d}\)	\(74\)
default	\(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )-2 b \tan \left (d x +c \right )}{b^{2}}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}}{d}\)	\(74\)
risch	\(-\frac {2 i \left (3 a \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a -5 b \right )}{3 d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{\sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}\)	\(446\)

[In]

int(sec(d*x+c)^6/(a+b*tan(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 339, normalized size of antiderivative = 4.40 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [-\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a b} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 4 \, {\left (a b^{2} - {\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a b^{3} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 2 \, {\left (a b^{2} - {\left (3 \, a^{2} b - 5 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, a b^{3} d \cos \left (d x + c\right )^{3}}\right ] \]

[In]

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a^2 - 2*a*b + b^2)*sqrt(-a*b)*cos(d*x + c)^3*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b
^2)*cos(d*x + c)^2 + 4*((a + b)*cos(d*x + c)^3 - b*cos(d*x + c))*sqrt(-a*b)*sin(d*x + c) + b^2)/((a^2 - 2*a*b
+ b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c)^2 + b^2)) - 4*(a*b^2 - (3*a^2*b - 5*a*b^2)*cos(d*x + c)^2)*
sin(d*x + c))/(a*b^3*d*cos(d*x + c)^3), -1/6*(3*(a^2 - 2*a*b + b^2)*sqrt(a*b)*arctan(1/2*((a + b)*cos(d*x + c)
^2 - b)*sqrt(a*b)/(a*b*cos(d*x + c)*sin(d*x + c)))*cos(d*x + c)^3 - 2*(a*b^2 - (3*a^2*b - 5*a*b^2)*cos(d*x + c
)^2)*sin(d*x + c))/(a*b^3*d*cos(d*x + c)^3)]

Sympy [F]

\[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {\sec ^{6}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]

[In]

integrate(sec(d*x+c)**6/(a+b*tan(d*x+c)**2),x)

[Out]

Integral(sec(c + d*x)**6/(a + b*tan(c + d*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b \tan \left (d x + c\right )^{3} - 3 \, {\left (a - 2 \, b\right )} \tan \left (d x + c\right )}{b^{2}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (a^{2} - 2 \, a b + b^{2}\right )}}{\sqrt {a b} b^{2}} + \frac {b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right ) + 6 \, b^{2} \tan \left (d x + c\right )}{b^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))*(a^2 - 2*a*b + b^2)/(sqrt(a*b)
*b^2) + (b^2*tan(d*x + c)^3 - 3*a*b*tan(d*x + c) + 6*b^2*tan(d*x + c))/b^3)/d

Mupad [B] (verification not implemented)

Time = 11.98 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a}{b^2}-\frac {2}{b}\right )}{d}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,\left (a^2-2\,a\,b+b^2\right )}\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,b^{5/2}\,d} \]

[In]

int(1/(cos(c + d*x)^6*(a + b*tan(c + d*x)^2)),x)

[Out]

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