Integrand size = 23, antiderivative size = 74 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tan (c+d x)}{d}+\frac {a (a+2 b) \tan ^3(c+d x)}{3 d}+\frac {b (2 a+b) \tan ^5(c+d x)}{5 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \] Output:
a^2*tan(d*x+c)/d+1/3*a*(a+2*b)*tan(d*x+c)^3/d+1/5*b*(2*a+b)*tan(d*x+c)^5/d +1/7*b^2*tan(d*x+c)^7/d
Time = 0.83 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\left (70 a^2-28 a b+6 b^2+\left (35 a^2-14 a b+3 b^2\right ) \sec ^2(c+d x)+6 (7 a-4 b) b \sec ^4(c+d x)+15 b^2 \sec ^6(c+d x)\right ) \tan (c+d x)}{105 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]
Output:
((70*a^2 - 28*a*b + 6*b^2 + (35*a^2 - 14*a*b + 3*b^2)*Sec[c + d*x]^2 + 6*( 7*a - 4*b)*b*Sec[c + d*x]^4 + 15*b^2*Sec[c + d*x]^6)*Tan[c + d*x])/(105*d)
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 290, 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.
\(\displaystyle \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^4 \left (a+b \tan (c+d x)^2\right )^2dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \frac {\int \left (b^2 \tan ^6(c+d x)+b (2 a+b) \tan ^4(c+d x)+a (a+2 b) \tan ^2(c+d x)+a^2\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \tan (c+d x)+\frac {1}{5} b (2 a+b) \tan ^5(c+d x)+\frac {1}{3} a (a+2 b) \tan ^3(c+d x)+\frac {1}{7} b^2 \tan ^7(c+d x)}{d}\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]
Output:
(a^2*Tan[c + d*x] + (a*(a + 2*b)*Tan[c + d*x]^3)/3 + (b*(2*a + b)*Tan[c + d*x]^5)/5 + (b^2*Tan[c + d*x]^7)/7)/d
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 9.72 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.50
method | result | size |
derivativedivides | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \sin \left (d x +c \right )^{5}}{35 \cos \left (d x +c \right )^{5}}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(111\) |
default | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \sin \left (d x +c \right )^{5}}{35 \cos \left (d x +c \right )^{5}}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(111\) |
risch | \(\frac {4 i \left (105 a^{2} {\mathrm e}^{10 i \left (d x +c \right )}-210 a b \,{\mathrm e}^{10 i \left (d x +c \right )}+105 b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+455 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-350 a b \,{\mathrm e}^{8 i \left (d x +c \right )}-105 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+770 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-140 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+210 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+630 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-84 \,{\mathrm e}^{4 i \left (d x +c \right )} a b -42 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+245 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-98 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+21 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+35 a^{2}-14 a b +3 b^{2}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\) | \(240\) |
Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(b^2*(1/7*sin(d*x+c)^5/cos(d*x+c)^7+2/35*sin(d*x+c)^5/cos(d*x+c)^5)+2* a*b*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)-a^2*(-2 /3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {{\left (2 \, {\left (35 \, a^{2} - 14 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + {\left (35 \, a^{2} - 14 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (7 \, a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")
Output:
1/105*(2*(35*a^2 - 14*a*b + 3*b^2)*cos(d*x + c)^6 + (35*a^2 - 14*a*b + 3*b ^2)*cos(d*x + c)^4 + 6*(7*a*b - 4*b^2)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c)/(d*cos(d*x + c)^7)
\[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*tan(d*x+c)**2)**2,x)
Output:
Integral((a + b*tan(c + d*x)**2)**2*sec(c + d*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 21 \, {\left (2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (a^{2} + 2 \, a b\right )} \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/105*(15*b^2*tan(d*x + c)^7 + 21*(2*a*b + b^2)*tan(d*x + c)^5 + 35*(a^2 + 2*a*b)*tan(d*x + c)^3 + 105*a^2*tan(d*x + c))/d
Time = 0.47 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 42 \, a b \tan \left (d x + c\right )^{5} + 21 \, b^{2} \tan \left (d x + c\right )^{5} + 35 \, a^{2} \tan \left (d x + c\right )^{3} + 70 \, a b \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/105*(15*b^2*tan(d*x + c)^7 + 42*a*b*tan(d*x + c)^5 + 21*b^2*tan(d*x + c) ^5 + 35*a^2*tan(d*x + c)^3 + 70*a*b*tan(d*x + c)^3 + 105*a^2*tan(d*x + c)) /d
Time = 8.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a+2\,b\right )}{3}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (2\,a+b\right )}{5}}{d} \] Input:
int((a + b*tan(c + d*x)^2)^2/cos(c + d*x)^4,x)
Output:
(a^2*tan(c + d*x) + (b^2*tan(c + d*x)^7)/7 + (a*tan(c + d*x)^3*(a + 2*b))/ 3 + (b*tan(c + d*x)^5*(2*a + b))/5)/d
Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.14 \[ \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\sin \left (d x +c \right ) \left (70 \sin \left (d x +c \right )^{6} a^{2}-28 \sin \left (d x +c \right )^{6} a b +6 \sin \left (d x +c \right )^{6} b^{2}-245 \sin \left (d x +c \right )^{4} a^{2}+98 \sin \left (d x +c \right )^{4} a b -21 \sin \left (d x +c \right )^{4} b^{2}+280 \sin \left (d x +c \right )^{2} a^{2}-70 \sin \left (d x +c \right )^{2} a b -105 a^{2}\right )}{105 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x)
Output:
(sin(c + d*x)*(70*sin(c + d*x)**6*a**2 - 28*sin(c + d*x)**6*a*b + 6*sin(c + d*x)**6*b**2 - 245*sin(c + d*x)**4*a**2 + 98*sin(c + d*x)**4*a*b - 21*si n(c + d*x)**4*b**2 + 280*sin(c + d*x)**2*a**2 - 70*sin(c + d*x)**2*a*b - 1 05*a**2))/(105*cos(c + d*x)*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin (c + d*x)**2 - 1))