\(\int \sec ^6(c+d x) (a+b \tan ^2(c+d x)) \, dx\) [431]

Optimal result

Integrand size = 21, antiderivative size = 68 \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {a \tan (c+d x)}{d}+\frac {(2 a+b) \tan ^3(c+d x)}{3 d}+\frac {(a+2 b) \tan ^5(c+d x)}{5 d}+\frac {b \tan ^7(c+d x)}{7 d} \]

[Out]

a*tan(d*x+c)/d+1/3*(2*a+b)*tan(d*x+c)^3/d+1/5*(a+2*b)*tan(d*x+c)^5/d+1/7*b*tan(d*x+c)^7/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3756, 380} \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {(a+2 b) \tan ^5(c+d x)}{5 d}+\frac {(2 a+b) \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \tan ^7(c+d x)}{7 d} \]

[In]

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

[Out]

(a*Tan[c + d*x])/d + ((2*a + b)*Tan[c + d*x]^3)/(3*d) + ((a + 2*b)*Tan[c + d*x]^5)/(5*d) + (b*Tan[c + d*x]^7)/
(7*d)

Rule 380

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

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 \left (1+x^2\right )^2 \left (a+b x^2\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (a+(2 a+b) x^2+(a+2 b) x^4+b x^6\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {a \tan (c+d x)}{d}+\frac {(2 a+b) \tan ^3(c+d x)}{3 d}+\frac {(a+2 b) \tan ^5(c+d x)}{5 d}+\frac {b \tan ^7(c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.10 \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {\tan (c+d x) \left (105 a-8 b-4 b \sec ^2(c+d x)-3 b \sec ^4(c+d x)+15 b \sec ^6(c+d x)+70 a \tan ^2(c+d x)+21 a \tan ^4(c+d x)\right )}{105 d} \]

[In]

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

[Out]

(Tan[c + d*x]*(105*a - 8*b - 4*b*Sec[c + d*x]^2 - 3*b*Sec[c + d*x]^4 + 15*b*Sec[c + d*x]^6 + 70*a*Tan[c + d*x]
^2 + 21*a*Tan[c + d*x]^4))/(105*d)

Maple [A] (verified)

Time = 4.83 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {\frac {b \tan \left (d x +c \right )^{7}}{7}+\frac {\left (a +2 b \right ) \tan \left (d x +c \right )^{5}}{5}+\frac {\left (2 a +b \right ) \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )}{d}\)	\(55\)
default	\(\frac {\frac {b \tan \left (d x +c \right )^{7}}{7}+\frac {\left (a +2 b \right ) \tan \left (d x +c \right )^{5}}{5}+\frac {\left (2 a +b \right ) \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )}{d}\)	\(55\)
risch	\(\frac {16 i \left (70 a \,{\mathrm e}^{8 i \left (d x +c \right )}-70 b \,{\mathrm e}^{8 i \left (d x +c \right )}+175 a \,{\mathrm e}^{6 i \left (d x +c \right )}+35 b \,{\mathrm e}^{6 i \left (d x +c \right )}+147 a \,{\mathrm e}^{4 i \left (d x +c \right )}-21 b \,{\mathrm e}^{4 i \left (d x +c \right )}+49 a \,{\mathrm e}^{2 i \left (d x +c \right )}-7 b \,{\mathrm e}^{2 i \left (d x +c \right )}+7 a -b \right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\)	\(123\)

[In]

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

[Out]

1/d*(1/7*b*tan(d*x+c)^7+1/5*(a+2*b)*tan(d*x+c)^5+1/3*(2*a+b)*tan(d*x+c)^3+a*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {{\left (8 \, {\left (7 \, a - b\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (7 \, a - b\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (7 \, a - b\right )} \cos \left (d x + c\right )^{2} + 15 \, b\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]

[In]

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

[Out]

1/105*(8*(7*a - b)*cos(d*x + c)^6 + 4*(7*a - b)*cos(d*x + c)^4 + 3*(7*a - b)*cos(d*x + c)^2 + 15*b)*sin(d*x +
c)/(d*cos(d*x + c)^7)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {15 \, b \tan \left (d x + c\right )^{7} + 21 \, {\left (a + 2 \, b\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (2 \, a + b\right )} \tan \left (d x + c\right )^{3} + 105 \, a \tan \left (d x + c\right )}{105 \, d} \]

[In]

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

[Out]

1/105*(15*b*tan(d*x + c)^7 + 21*(a + 2*b)*tan(d*x + c)^5 + 35*(2*a + b)*tan(d*x + c)^3 + 105*a*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {15 \, b \tan \left (d x + c\right )^{7} + 21 \, a \tan \left (d x + c\right )^{5} + 42 \, b \tan \left (d x + c\right )^{5} + 70 \, a \tan \left (d x + c\right )^{3} + 35 \, b \tan \left (d x + c\right )^{3} + 105 \, a \tan \left (d x + c\right )}{105 \, d} \]

[In]

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

[Out]

1/105*(15*b*tan(d*x + c)^7 + 21*a*tan(d*x + c)^5 + 42*b*tan(d*x + c)^5 + 70*a*tan(d*x + c)^3 + 35*b*tan(d*x +
c)^3 + 105*a*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 12.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \sec ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+\left (\frac {a}{5}+\frac {2\,b}{5}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (\frac {2\,a}{3}+\frac {b}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+a\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

[In]

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

[Out]

(tan(c + d*x)^3*((2*a)/3 + b/3) + tan(c + d*x)^5*(a/5 + (2*b)/5) + a*tan(c + d*x) + (b*tan(c + d*x)^7)/7)/d