Integrand size = 21, antiderivative size = 88 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^{1+n}}{b^3 d (1+n)}-\frac {2 a (a+b \tan (c+d x))^{2+n}}{b^3 d (2+n)}+\frac {(a+b \tan (c+d x))^{3+n}}{b^3 d (3+n)} \] Output:
(a^2+b^2)*(a+b*tan(d*x+c))^(1+n)/b^3/d/(1+n)-2*a*(a+b*tan(d*x+c))^(2+n)/b^ 3/d/(2+n)+(a+b*tan(d*x+c))^(3+n)/b^3/d/(3+n)
Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (\frac {a^2+b^2}{1+n}-\frac {2 a (a+b \tan (c+d x))}{2+n}+\frac {(a+b \tan (c+d x))^2}{3+n}\right )}{b^3 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^n,x]
Output:
((a + b*Tan[c + d*x])^(1 + n)*((a^2 + b^2)/(1 + n) - (2*a*(a + b*Tan[c + d *x]))/(2 + n) + (a + b*Tan[c + d*x])^2/(3 + n)))/(b^3*d)
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3987, 27, 476, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^4 (a+b \tan (c+d x))^ndx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^n \left (\tan ^2(c+d x) b^2+b^2\right )}{b^2}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a+b \tan (c+d x))^n \left (\tan ^2(c+d x) b^2+b^2\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (\left (a^2+b^2\right ) (a+b \tan (c+d x))^n-2 a (a+b \tan (c+d x))^{n+1}+(a+b \tan (c+d x))^{n+2}\right )d(b \tan (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\left (a^2+b^2\right ) (a+b \tan (c+d x))^{n+1}}{n+1}-\frac {2 a (a+b \tan (c+d x))^{n+2}}{n+2}+\frac {(a+b \tan (c+d x))^{n+3}}{n+3}}{b^3 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x])^n,x]
Output:
(((a^2 + b^2)*(a + b*Tan[c + d*x])^(1 + n))/(1 + n) - (2*a*(a + b*Tan[c + d*x])^(2 + n))/(2 + n) + (a + b*Tan[c + d*x])^(3 + n)/(3 + n))/(b^3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(203\) vs. \(2(88)=176\).
Time = 181.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.32
method | result | size |
derivativedivides | \(\frac {\tan \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {a \left (b^{2} n^{2}+5 b^{2} n +2 a^{2}+6 b^{2}\right ) {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{3} d \left (n^{3}+6 n^{2}+11 n +6\right )}+\frac {a n \tan \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b d \left (n^{2}+5 n +6\right )}-\frac {\left (-b^{2} n^{2}+2 a^{2} n -5 b^{2} n -6 b^{2}\right ) \tan \left (d x +c \right ) {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{2} \left (n^{3}+6 n^{2}+11 n +6\right ) d}\) | \(204\) |
default | \(\frac {\tan \left (d x +c \right )^{3} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {a \left (b^{2} n^{2}+5 b^{2} n +2 a^{2}+6 b^{2}\right ) {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{3} d \left (n^{3}+6 n^{2}+11 n +6\right )}+\frac {a n \tan \left (d x +c \right )^{2} {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b d \left (n^{2}+5 n +6\right )}-\frac {\left (-b^{2} n^{2}+2 a^{2} n -5 b^{2} n -6 b^{2}\right ) \tan \left (d x +c \right ) {\mathrm e}^{n \ln \left (a +b \tan \left (d x +c \right )\right )}}{b^{2} \left (n^{3}+6 n^{2}+11 n +6\right ) d}\) | \(204\) |
Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c))^n,x,method=_RETURNVERBOSE)
Output:
1/d/(3+n)*tan(d*x+c)^3*exp(n*ln(a+b*tan(d*x+c)))+a*(b^2*n^2+5*b^2*n+2*a^2+ 6*b^2)/b^3/d/(n^3+6*n^2+11*n+6)*exp(n*ln(a+b*tan(d*x+c)))+a*n/b/d/(n^2+5*n +6)*tan(d*x+c)^2*exp(n*ln(a+b*tan(d*x+c)))-(-b^2*n^2+2*a^2*n-5*b^2*n-6*b^2 )/b^2/(n^3+6*n^2+11*n+6)/d*tan(d*x+c)*exp(n*ln(a+b*tan(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.00 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {{\left (2 \, {\left (2 \, a b^{2} n + a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a b^{2} n^{2} + a b^{2} n\right )} \cos \left (d x + c\right ) + {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3} + 2 \, {\left (2 \, b^{3} - {\left (a^{2} b - b^{3}\right )} n\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \left (\frac {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )^{n}}{{\left (b^{3} d n^{3} + 6 \, b^{3} d n^{2} + 11 \, b^{3} d n + 6 \, b^{3} d\right )} \cos \left (d x + c\right )^{3}} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="fricas")
Output:
(2*(2*a*b^2*n + a^3 + 3*a*b^2)*cos(d*x + c)^3 + (a*b^2*n^2 + a*b^2*n)*cos( d*x + c) + (b^3*n^2 + 3*b^3*n + 2*b^3 + 2*(2*b^3 - (a^2*b - b^3)*n)*cos(d* x + c)^2)*sin(d*x + c))*((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))^n /((b^3*d*n^3 + 6*b^3*d*n^2 + 11*b^3*d*n + 6*b^3*d)*cos(d*x + c)^3)
\[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*tan(d*x+c))**n,x)
Output:
Integral((a + b*tan(c + d*x))**n*sec(c + d*x)**4, x)
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32 \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{n + 1}}{b {\left (n + 1\right )}} + \frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} \tan \left (d x + c\right )^{3} + {\left (n^{2} + n\right )} a b^{2} \tan \left (d x + c\right )^{2} - 2 \, a^{2} b n \tan \left (d x + c\right ) + 2 \, a^{3}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}}}{d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="maxima")
Output:
((b*tan(d*x + c) + a)^(n + 1)/(b*(n + 1)) + ((n^2 + 3*n + 2)*b^3*tan(d*x + c)^3 + (n^2 + n)*a*b^2*tan(d*x + c)^2 - 2*a^2*b*n*tan(d*x + c) + 2*a^3)*( b*tan(d*x + c) + a)^n/((n^3 + 6*n^2 + 11*n + 6)*b^3))/d
\[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="giac")
Output:
integrate((b*tan(d*x + c) + a)^n*sec(d*x + c)^4, x)
Timed out. \[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:
int((a + b*tan(c + d*x))^n/cos(c + d*x)^4,x)
Output:
int((a + b*tan(c + d*x))^n/cos(c + d*x)^4, x)
\[ \int \sec ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (\tan \left (d x +c \right ) b +a \right )^{n} \sec \left (d x +c \right )^{4}d x \] Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c))^n,x)
Output:
int((tan(c + d*x)*b + a)**n*sec(c + d*x)**4,x)