Integrand size = 21, antiderivative size = 99 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (4 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a b \sec ^3(c+d x)}{12 d}+\frac {\left (4 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d} \]
1/8*(4*a^2-b^2)*arctanh(sin(d*x+c))/d+5/12*a*b*sec(d*x+c)^3/d+1/8*(4*a^2-b ^2)*sec(d*x+c)*tan(d*x+c)/d+1/4*b*sec(d*x+c)^3*(a+b*tan(d*x+c))/d
Time = 0.06 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
(a^2*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*ArcTanh[Sin[c + d*x]])/(8*d) + (2 *a*b*Sec[c + d*x]^3)/(3*d) + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) - (b^2* Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.40 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3991, 27, 3042, 3086, 15, 4159, 298, 215, 219}
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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^3 (a+b \tan (c+d x))^2dx\) |
| \(\Big \downarrow \) 3991 |
| \(\displaystyle \int \sec ^3(c+d x) \left (a^2+b^2 \tan ^2(c+d x)\right )dx+\int 2 a b \sec ^3(c+d x) \tan (c+d x)dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \sec ^3(c+d x) \left (a^2+b^2 \tan ^2(c+d x)\right )dx+2 a b \int \sec ^3(c+d x) \tan (c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^3 \left (a^2+b^2 \tan (c+d x)^2\right )dx+2 a b \int \sec (c+d x)^3 \tan (c+d x)dx\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \int \sec (c+d x)^3 \left (a^2+b^2 \tan (c+d x)^2\right )dx+\frac {2 a b \int \sec ^2(c+d x)d\sec (c+d x)}{d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \int \sec (c+d x)^3 \left (a^2+b^2 \tan (c+d x)^2\right )dx+\frac {2 a b \sec ^3(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4159 |
| \(\displaystyle \frac {\int \frac {a^2-\left (a^2-b^2\right ) \sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^3}d\sin (c+d x)}{d}+\frac {2 a b \sec ^3(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {1}{4} \left (4 a^2-b^2\right ) \int \frac {1}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)+\frac {b^2 \sin (c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}}{d}+\frac {2 a b \sec ^3(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{4} \left (4 a^2-b^2\right ) \left (\frac {1}{2} \int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)+\frac {\sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )+\frac {b^2 \sin (c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}}{d}+\frac {2 a b \sec ^3(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{4} \left (4 a^2-b^2\right ) \left (\frac {1}{2} \text {arctanh}(\sin (c+d x))+\frac {\sin (c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}\right )+\frac {b^2 \sin (c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}}{d}+\frac {2 a b \sec ^3(c+d x)}{3 d}\) |
(2*a*b*Sec[c + d*x]^3)/(3*d) + ((b^2*Sin[c + d*x])/(4*(1 - Sin[c + d*x]^2) ^2) + ((4*a^2 - b^2)*(ArcTanh[Sin[c + d*x]]/2 + Sin[c + d*x]/(2*(1 - Sin[c + d*x]^2))))/4)/d
3.6.25.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Module[{k}, Int[Sec[e + f*x]^m*Sum[Binomial[n, 2*k]*a^(n - 2*k)*b^(2*k)*Tan[e + f*x]^(2*k), {k, 0, n/2}], x] + Int[Sec[e + f*x]^m*Tan [e + f*x]*Sum[Binomial[n, 2*k + 1]*a^(n - 2*k - 1)*b^(2*k + 1)*Tan[e + f*x] ^(2*k), {k, 0, (n - 1)/2}], x]] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 4.72 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(118\) |
| default | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(118\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (12 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+21 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+64 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-21 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+64 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2}+3 b^{2}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}\) | \(255\) |
1/d*(a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+2/3*a*b /cos(d*x+c)^3+b^2*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+ c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 32 \, a b \cos \left (d x + c\right ) + 6 \, {\left ({\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(3*(4*a^2 - b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*a^2 - b^ 2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 32*a*b*cos(d*x + c) + 6*((4*a^2 - b^2)*cos(d*x + c)^2 + 2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {32 \, a b}{\cos \left (d x + c\right )^{3}}}{48 \, d} \]
1/48*(3*b^2*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*a^2*(2* sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 32*a*b/cos(d*x + c)^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (91) = 182\).
Time = 0.51 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.52 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {3 \, {\left (4 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16 \, a b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(4*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2 - b^2) *log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(12*a^2*tan(1/2*d*x + 1/2*c)^7 + 3 *b^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*b*tan(1/2*d*x + 1/2*c)^6 - 12*a^2*tan(1 /2*d*x + 1/2*c)^5 + 21*b^2*tan(1/2*d*x + 1/2*c)^5 + 48*a*b*tan(1/2*d*x + 1 /2*c)^4 - 12*a^2*tan(1/2*d*x + 1/2*c)^3 + 21*b^2*tan(1/2*d*x + 1/2*c)^3 - 16*a*b*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/ 2*d*x + 1/2*c) + 16*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 7.25 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.18 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+\frac {b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {7\,b^2}{4}-a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {7\,b^2}{4}-a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\left (a^2+\frac {b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4\,a\,b}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {b^2}{4}\right )}{d} \]
((4*a*b)/3 + tan(c/2 + (d*x)/2)*(a^2 + b^2/4) + tan(c/2 + (d*x)/2)^7*(a^2 + b^2/4) - tan(c/2 + (d*x)/2)^3*(a^2 - (7*b^2)/4) - tan(c/2 + (d*x)/2)^5*( a^2 - (7*b^2)/4) - (4*a*b*tan(c/2 + (d*x)/2)^2)/3 + 4*a*b*tan(c/2 + (d*x)/ 2)^4 - 4*a*b*tan(c/2 + (d*x)/2)^6)/(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)) + (atan h(tan(c/2 + (d*x)/2))*(a^2 - b^2/4))/d