\(\int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx\) [533]

Optimal result

Integrand size = 21, antiderivative size = 75 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^4}{4 b^3 d}-\frac {2 a (a+b \tan (c+d x))^5}{5 b^3 d}+\frac {(a+b \tan (c+d x))^6}{6 b^3 d} \]

[Out]

1/4*(a^2+b^2)*(a+b*tan(d*x+c))^4/b^3/d-2/5*a*(a+b*tan(d*x+c))^5/b^3/d+1/6*(a+b*tan(d*x+c))^6/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} \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^4}{4 b^3 d}+\frac {(a+b \tan (c+d x))^6}{6 b^3 d}-\frac {2 a (a+b \tan (c+d x))^5}{5 b^3 d} \]

[In]

Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]

[Out]

((a^2 + b^2)*(a + b*Tan[c + d*x])^4)/(4*b^3*d) - (2*a*(a + b*Tan[c + d*x])^5)/(5*b^3*d) + (a + b*Tan[c + d*x])
^6/(6*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)^3 \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)^3}{b^2}-\frac {2 a (a+x)^4}{b^2}+\frac {(a+x)^5}{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))^4}{4 b^3 d}-\frac {2 a (a+b \tan (c+d x))^5}{5 b^3 d}+\frac {(a+b \tan (c+d x))^6}{6 b^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {(a+b \tan (c+d x))^4 \left (a^2+15 b^2-4 a b \tan (c+d x)+10 b^2 \tan ^2(c+d x)\right )}{60 b^3 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^3,x]

[Out]

((a + b*Tan[c + d*x])^4*(a^2 + 15*b^2 - 4*a*b*Tan[c + d*x] + 10*b^2*Tan[c + d*x]^2))/(60*b^3*d)

Maple [A] (verified)

Time = 16.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.69

method	result	size
derivativedivides	\(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\)	\(127\)
default	\(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\)	\(127\)
risch	\(-\frac {4 \left (-15 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-50 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+30 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-90 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-10 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-60 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-30 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+18 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 i a^{3}+3 i a \,b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(228\)

[In]

int(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-a^3*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+3/4*a^2*b/cos(d*x+c)^4+3*a*b^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/1
5*sin(d*x+c)^3/cos(d*x+c)^3)+b^3*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.40 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {10 \, b^{3} + 15 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 9 \, a b^{2} \cos \left (d x + c\right ) + {\left (5 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{6}} \]

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(10*b^3 + 15*(3*a^2*b - b^3)*cos(d*x + c)^2 + 4*(2*(5*a^3 - 3*a*b^2)*cos(d*x + c)^5 + 9*a*b^2*cos(d*x + c
) + (5*a^3 - 3*a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^6)

Sympy [F]

\[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**4*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.31 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 15 \, {\left (3 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 60 \, a^{3} \tan \left (d x + c\right ) + 20 \, {\left (a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(10*b^3*tan(d*x + c)^6 + 36*a*b^2*tan(d*x + c)^5 + 90*a^2*b*tan(d*x + c)^2 + 15*(3*a^2*b + b^3)*tan(d*x +
c)^4 + 60*a^3*tan(d*x + c) + 20*(a^3 + 3*a*b^2)*tan(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.75 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {10 \, b^{3} \tan \left (d x + c\right )^{6} + 36 \, a b^{2} \tan \left (d x + c\right )^{5} + 45 \, a^{2} b \tan \left (d x + c\right )^{4} + 15 \, b^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} + 60 \, a b^{2} \tan \left (d x + c\right )^{3} + 90 \, a^{2} b \tan \left (d x + c\right )^{2} + 60 \, a^{3} \tan \left (d x + c\right )}{60 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(10*b^3*tan(d*x + c)^6 + 36*a*b^2*tan(d*x + c)^5 + 45*a^2*b*tan(d*x + c)^4 + 15*b^3*tan(d*x + c)^4 + 20*a
^3*tan(d*x + c)^3 + 60*a*b^2*tan(d*x + c)^3 + 90*a^2*b*tan(d*x + c)^2 + 60*a^3*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{4}+\frac {b^3}{4}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}}{d} \]

[In]

int((a + b*tan(c + d*x))^3/cos(c + d*x)^4,x)

[Out]

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