\(\int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx\) [519]

Optimal result

Integrand size = 21, antiderivative size = 75 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^3}{3 b^3 d}-\frac {a (a+b \tan (c+d x))^4}{2 b^3 d}+\frac {(a+b \tan (c+d x))^5}{5 b^3 d} \]

[Out]

1/3*(a^2+b^2)*(a+b*tan(d*x+c))^3/b^3/d-1/2*a*(a+b*tan(d*x+c))^4/b^3/d+1/5*(a+b*tan(d*x+c))^5/b^3/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 75, 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 = {3587, 711} \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^3}{3 b^3 d}+\frac {(a+b \tan (c+d x))^5}{5 b^3 d}-\frac {a (a+b \tan (c+d x))^4}{2 b^3 d} \]

[In]

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

[Out]

((a^2 + b^2)*(a + b*Tan[c + d*x])^3)/(3*b^3*d) - (a*(a + b*Tan[c + d*x])^4)/(2*b^3*d) + (a + b*Tan[c + d*x])^5
/(5*b^3*d)

Rule 711

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {(a+b \tan (c+d x))^3 \left (a^2+10 b^2-3 a b \tan (c+d x)+6 b^2 \tan ^2(c+d x)\right )}{30 b^3 d} \]

[In]

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

[Out]

((a + b*Tan[c + d*x])^3*(a^2 + 10*b^2 - 3*a*b*Tan[c + d*x] + 6*b^2*Tan[c + d*x]^2))/(30*b^3*d)

Maple [A] (verified)

Time = 7.98 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a b}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )}{d}\)	\(82\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a b}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )}{d}\)	\(82\)
risch	\(\frac {4 i \left (-30 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-15 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-30 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+35 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+25 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2}-b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(143\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \, a b \cos \left (d x + c\right ) + 2 \, {\left (2 \, {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/30*(15*a*b*cos(d*x + c) + 2*(2*(5*a^2 - b^2)*cos(d*x + c)^4 + (5*a^2 - b^2)*cos(d*x + c)^2 + 3*b^2)*sin(d*x
+ c))/(d*cos(d*x + c)^5)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} + 30 \, a b \tan \left (d x + c\right )^{2} + 10 \, {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 + 30*a*b*tan(d*x + c)^2 + 10*(a^2 + b^2)*tan(d*x + c)^3 + 3
0*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{3} + 10 \, b^{2} \tan \left (d x + c\right )^{3} + 30 \, a b \tan \left (d x + c\right )^{2} + 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 + 10*a^2*tan(d*x + c)^3 + 10*b^2*tan(d*x + c)^3 + 30*a*b*ta
n(d*x + c)^2 + 30*a^2*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 4.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a^2}{3}+\frac {b^2}{3}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2}}{d} \]

[In]

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

[Out]

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