\(\int \sec ^2(c+d x) (a+b \tan ^2(c+d x)) \, dx\) [433]

Optimal result

Integrand size = 21, antiderivative size = 28 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {a \tan (c+d x)}{d}+\frac {b \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3756} \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {a \tan (c+d x)}{d}+\frac {b \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

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 (a+b x^2\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {a \tan (c+d x)}{d}+\frac {b \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {a \tan (c+d x)}{d}+\frac {b \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {b \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )}{d}\)	\(25\)
default	\(\frac {\frac {b \tan \left (d x +c \right )^{3}}{3}+a \tan \left (d x +c \right )}{d}\)	\(25\)
risch	\(-\frac {2 i \left (-3 a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 a +b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {{\left ({\left (3 \, a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\begin {cases} \frac {a \tan {\left (c + d x \right )} + \frac {b \tan ^{3}{\left (c + d x \right )}}{3}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan ^{2}{\left (c \right )}\right ) \sec ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((a*tan(c + d*x) + b*tan(c + d*x)**3/3)/d, Ne(d, 0)), (x*(a + b*tan(c)**2)*sec(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {b \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )}{3 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {b \tan \left (d x + c\right )^{3} + 3 \, a \tan \left (d x + c\right )}{3 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2+3\,a\right )}{3\,d} \]

[In]

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

[Out]

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