\(\int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx\) [518]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a b \sec ^6(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

1/3*a*b*sec(d*x+c)^6/d+a^2*tan(d*x+c)/d+1/3*(2*a^2+b^2)*tan(d*x+c)^3/d+1/5*(a^2+2*b^2)*tan(d*x+c)^5/d+1/7*b^2*
tan(d*x+c)^7/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3587, 710, 1824} \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \sec ^6(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[In]

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

[Out]

(a*b*Sec[c + d*x]^6)/(3*d) + (a^2*Tan[c + d*x])/d + ((2*a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + ((a^2 + 2*b^2)*Tan[
c + d*x]^5)/(5*d) + (b^2*Tan[c + d*x]^7)/(7*d)

Rule 710

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

Rule 1824

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 3587

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

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

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\tan (c+d x) \left (105 a^2+105 a b \tan (c+d x)+35 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+105 a b \tan ^3(c+d x)+21 \left (a^2+2 b^2\right ) \tan ^4(c+d x)+35 a b \tan ^5(c+d x)+15 b^2 \tan ^6(c+d x)\right )}{105 d} \]

[In]

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

[Out]

(Tan[c + d*x]*(105*a^2 + 105*a*b*Tan[c + d*x] + 35*(2*a^2 + b^2)*Tan[c + d*x]^2 + 105*a*b*Tan[c + d*x]^3 + 21*
(a^2 + 2*b^2)*Tan[c + d*x]^4 + 35*a*b*Tan[c + d*x]^5 + 15*b^2*Tan[c + d*x]^6))/(105*d)

Maple [A] (verified)

Time = 28.99 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {a b}{3 \cos \left (d x +c \right )^{6}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )}{d}\)	\(110\)
default	\(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {a b}{3 \cos \left (d x +c \right )^{6}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )}{d}\)	\(110\)
risch	\(\frac {16 i \left (-140 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}+70 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-70 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-140 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+175 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+35 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+147 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-21 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+49 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+7 a^{2}-b^{2}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\)	\(171\)

[In]

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

[Out]

1/d*(-a^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/3*a*b/cos(d*x+c)^6+b^2*(1/7*sin(d*x+c)^3/cos
(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {35 \, a b \cos \left (d x + c\right ) + {\left (8 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]

[In]

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

[Out]

1/105*(35*a*b*cos(d*x + c) + (8*(7*a^2 - b^2)*cos(d*x + c)^6 + 4*(7*a^2 - b^2)*cos(d*x + c)^4 + 3*(7*a^2 - b^2
)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c))/(d*cos(d*x + c)^7)

Sympy [F]

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

[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 = 104, normalized size of antiderivative = 1.07 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 105 \, a b \tan \left (d x + c\right )^{4} + 21 \, {\left (a^{2} + 2 \, b^{2}\right )} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{2} + 35 \, {\left (2 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]

[In]

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

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 105*a*b*tan(d*x + c)^4 + 21*(a^2 + 2*b^2)*tan(d*x + c)^
5 + 105*a*b*tan(d*x + c)^2 + 35*(2*a^2 + b^2)*tan(d*x + c)^3 + 105*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} + 42 \, b^{2} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{4} + 70 \, a^{2} \tan \left (d x + c\right )^{3} + 35 \, b^{2} \tan \left (d x + c\right )^{3} + 105 \, a b \tan \left (d x + c\right )^{2} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]

[In]

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

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 + 42*b^2*tan(d*x + c)^5 + 105*a*b
*tan(d*x + c)^4 + 70*a^2*tan(d*x + c)^3 + 35*b^2*tan(d*x + c)^3 + 105*a*b*tan(d*x + c)^2 + 105*a^2*tan(d*x + c
))/d

Mupad [B] (verification not implemented)

Time = 4.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.05 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}+\frac {b^2}{3}\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {a^2}{5}+\frac {2\,b^2}{5}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4+\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^6}{3}}{d} \]

[In]

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

[Out]

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