\(\int \sec ^4(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\) [445]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tan (c+d x)}{d}+\frac {a (a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b (2 a+b) \tan ^5(c+d x)}{5 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3756, 380} \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tan (c+d x)}{d}+\frac {b (2 a+b) \tan ^5(c+d x)}{5 d}+\frac {a (a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]

[In]

Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\left (70 a^2-28 a b+6 b^2+\left (35 a^2-14 a b+3 b^2\right ) \sec ^2(c+d x)+6 (7 a-4 b) b \sec ^4(c+d x)+15 b^2 \sec ^6(c+d x)\right ) \tan (c+d x)}{105 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((70*a^2 - 28*a*b + 6*b^2 + (35*a^2 - 14*a*b + 3*b^2)*Sec[c + d*x]^2 + 6*(7*a - 4*b)*b*Sec[c + d*x]^4 + 15*b^2
*Sec[c + d*x]^6)*Tan[c + d*x])/(105*d)

Maple [A] (verified)

Time = 3.60 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \sin \left (d x +c \right )^{5}}{35 \cos \left (d x +c \right )^{5}}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(111\)
default	\(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \sin \left (d x +c \right )^{5}}{35 \cos \left (d x +c \right )^{5}}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(111\)
risch	\(\frac {4 i \left (105 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}-210 a b \,{\mathrm e}^{10 i \left (d x +c \right )}+105 b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+455 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-350 a b \,{\mathrm e}^{8 i \left (d x +c \right )}-105 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+770 \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}-140 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+210 \,{\mathrm e}^{6 i \left (d x +c \right )} b^{2}+630 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-84 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-42 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+245 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-98 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{2}+35 a^{2}-14 a b +3 b^{2}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\)	\(240\)

[In]

int(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {{\left (2 \, {\left (35 \, a^{2} - 14 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + {\left (35 \, a^{2} - 14 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (7 \, a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/105*(2*(35*a^2 - 14*a*b + 3*b^2)*cos(d*x + c)^6 + (35*a^2 - 14*a*b + 3*b^2)*cos(d*x + c)^4 + 6*(7*a*b - 4*b^
2)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c)/(d*cos(d*x + c)^7)

Sympy [F]

\[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**4*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tan(c + d*x)**2)**2*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 21 \, {\left (2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (a^{2} + 2 \, a b\right )} \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 42 \, a b \tan \left (d x + c\right )^{5} + 21 \, b^{2} \tan \left (d x + c\right )^{5} + 35 \, a^{2} \tan \left (d x + c\right )^{3} + 70 \, a b \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a+2\,b\right )}{3}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (2\,a+b\right )}{5}}{d} \]

[In]

int((a + b*tan(c + d*x)^2)^2/cos(c + d*x)^4,x)

[Out]

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