Integrand size = 21, antiderivative size = 87 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2 \log (\cos (c+d x))}{d}-\frac {2 a b \sec (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}+\frac {b^2 \sec ^4(c+d x)}{4 d} \] Output:
a^2*ln(cos(d*x+c))/d-2*a*b*sec(d*x+c)/d+1/2*(a^2-b^2)*sec(d*x+c)^2/d+2/3*a *b*sec(d*x+c)^3/d+1/4*b^2*sec(d*x+c)^4/d
Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 a^2 \log (\cos (c+d x))-24 a b \sec (c+d x)+6 \left (a^2-b^2\right ) \sec ^2(c+d x)+8 a b \sec ^3(c+d x)+3 b^2 \sec ^4(c+d x)}{12 d} \] Input:
Integrate[(a + b*Sec[c + d*x])^2*Tan[c + d*x]^3,x]
Output:
(12*a^2*Log[Cos[c + d*x]] - 24*a*b*Sec[c + d*x] + 6*(a^2 - b^2)*Sec[c + d* x]^2 + 8*a*b*Sec[c + d*x]^3 + 3*b^2*Sec[c + d*x]^4)/(12*d)
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 25, 4373, 522, 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.
\(\displaystyle \int \tan ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -\frac {\int \frac {\cos (c+d x) (a+b \sec (c+d x))^2 \left (b^2-b^2 \sec ^2(c+d x)\right )}{b}d(b \sec (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (-b^3 \sec ^3(c+d x)-2 a b^2 \sec ^2(c+d x)-b \left (a^2-b^2\right ) \sec (c+d x)+2 a b^2+a^2 b \cos (c+d x)\right )d(b \sec (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{2} b^2 \left (a^2-b^2\right ) \sec ^2(c+d x)+a^2 b^2 \log (b \sec (c+d x))-\frac {2}{3} a b^3 \sec ^3(c+d x)+2 a b^3 \sec (c+d x)-\frac {1}{4} b^4 \sec ^4(c+d x)}{b^2 d}\) |
Input:
Int[(a + b*Sec[c + d*x])^2*Tan[c + d*x]^3,x]
Output:
-((a^2*b^2*Log[b*Sec[c + d*x]] + 2*a*b^3*Sec[c + d*x] - (b^2*(a^2 - b^2)*S ec[c + d*x]^2)/2 - (2*a*b^3*Sec[c + d*x]^3)/3 - (b^4*Sec[c + d*x]^4)/4)/(b ^2*d))
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.67 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86
method | result | size |
parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )^{4}}{4 d}+\frac {2 a b \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}\) | \(75\) |
derivativedivides | \(\frac {\frac {b^{2} \sec \left (d x +c \right )^{4}}{4}+\frac {2 a b \sec \left (d x +c \right )^{3}}{3}+\frac {a^{2} \sec \left (d x +c \right )^{2}}{2}-\frac {\sec \left (d x +c \right )^{2} b^{2}}{2}-2 a b \sec \left (d x +c \right )-a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(79\) |
default | \(\frac {\frac {b^{2} \sec \left (d x +c \right )^{4}}{4}+\frac {2 a b \sec \left (d x +c \right )^{3}}{3}+\frac {a^{2} \sec \left (d x +c \right )^{2}}{2}-\frac {\sec \left (d x +c \right )^{2} b^{2}}{2}-2 a b \sec \left (d x +c \right )-a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(79\) |
risch | \(-i a^{2} x -\frac {2 i a^{2} c}{d}-\frac {2 \left (6 a b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+10 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+10 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(179\) |
Input:
int((a+b*sec(d*x+c))^2*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
a^2/d*(1/2*tan(d*x+c)^2-1/2*ln(1+tan(d*x+c)^2))+1/4*b^2*tan(d*x+c)^4/d+2*a *b/d*(1/3*sec(d*x+c)^3-sec(d*x+c))
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) - 24 \, a b \cos \left (d x + c\right )^{3} + 8 \, a b \cos \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}}{12 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*sec(d*x+c))^2*tan(d*x+c)^3,x, algorithm="fricas")
Output:
1/12*(12*a^2*cos(d*x + c)^4*log(-cos(d*x + c)) - 24*a*b*cos(d*x + c)^3 + 8 *a*b*cos(d*x + c) + 6*(a^2 - b^2)*cos(d*x + c)^2 + 3*b^2)/(d*cos(d*x + c)^ 4)
Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\begin {cases} - \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {2 a b \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{3 d} - \frac {4 a b \sec {\left (c + d x \right )}}{3 d} + \frac {b^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \sec ^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\left (c \right )}\right )^{2} \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sec(d*x+c))**2*tan(d*x+c)**3,x)
Output:
Piecewise((-a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*tan(c + d*x)**2/(2* d) + 2*a*b*tan(c + d*x)**2*sec(c + d*x)/(3*d) - 4*a*b*sec(c + d*x)/(3*d) + b**2*tan(c + d*x)**2*sec(c + d*x)**2/(4*d) - b**2*sec(c + d*x)**2/(4*d), Ne(d, 0)), (x*(a + b*sec(c))**2*tan(c)**3, True))
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {24 \, a b \cos \left (d x + c\right )^{3} - 8 \, a b \cos \left (d x + c\right ) - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, b^{2}}{\cos \left (d x + c\right )^{4}}}{12 \, d} \] Input:
integrate((a+b*sec(d*x+c))^2*tan(d*x+c)^3,x, algorithm="maxima")
Output:
1/12*(12*a^2*log(cos(d*x + c)) - (24*a*b*cos(d*x + c)^3 - 8*a*b*cos(d*x + c) - 6*(a^2 - b^2)*cos(d*x + c)^2 - 3*b^2)/cos(d*x + c)^4)/d
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 \, a^{2} \log \left ({\left | \cos \left (d x + c\right ) \right |}\right ) - \frac {24 \, a b \cos \left (d x + c\right )^{3} - 8 \, a b \cos \left (d x + c\right ) - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, b^{2}}{\cos \left (d x + c\right )^{4}}}{12 \, d} \] Input:
integrate((a+b*sec(d*x+c))^2*tan(d*x+c)^3,x, algorithm="giac")
Output:
1/12*(12*a^2*log(abs(cos(d*x + c))) - (24*a*b*cos(d*x + c)^3 - 8*a*b*cos(d *x + c) - 6*(a^2 - b^2)*cos(d*x + c)^2 - 3*b^2)/cos(d*x + c)^4)/d
Time = 12.90 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.74 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {\frac {8\,a\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+\frac {32\,b\,a}{3}\right )-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+8\,a\,b-4\,b^2\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \] Input:
int(tan(c + d*x)^3*(a + b/cos(c + d*x))^2,x)
Output:
- ((8*a*b)/3 - tan(c/2 + (d*x)/2)^2*((32*a*b)/3 + 2*a^2) - 2*a^2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^4*(8*a*b + 4*a^2 - 4*b^2))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (2*a^2*atanh(tan(c/2 + (d*x)/2)^2))/d
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.11 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{2}+3 \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2} b^{2}-3 \sec \left (d x +c \right )^{2} b^{2}+8 \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2} a b -16 \sec \left (d x +c \right ) a b +6 \tan \left (d x +c \right )^{2} a^{2}}{12 d} \] Input:
int((a+b*sec(d*x+c))^2*tan(d*x+c)^3,x)
Output:
( - 6*log(tan(c + d*x)**2 + 1)*a**2 + 3*sec(c + d*x)**2*tan(c + d*x)**2*b* *2 - 3*sec(c + d*x)**2*b**2 + 8*sec(c + d*x)*tan(c + d*x)**2*a*b - 16*sec( c + d*x)*a*b + 6*tan(c + d*x)**2*a**2)/(12*d)