Integrand size = 15, antiderivative size = 67 \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\frac {\cos (a-c) \sec ^3(c+b x)}{3 b}+\frac {\text {arctanh}(\sin (c+b x)) \sin (a-c)}{2 b}+\frac {\sec (c+b x) \sin (a-c) \tan (c+b x)}{2 b} \] Output:
1/3*cos(a-c)*sec(b*x+c)^3/b+1/2*arctanh(sin(b*x+c))*sin(a-c)/b+1/2*sec(b*x +c)*sin(a-c)*tan(b*x+c)/b
Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\frac {12 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \sin (a-c)+\sec ^3(c+b x) (4 \cos (a-c)+3 \sin (a-c) \sin (2 (c+b x)))}{12 b} \] Input:
Integrate[Sec[c + b*x]^4*Sin[a + b*x],x]
Output:
(12*ArcTanh[Sin[c] + Cos[c]*Tan[(b*x)/2]]*Sin[a - c] + Sec[c + b*x]^3*(4*C os[a - c] + 3*Sin[a - c]*Sin[2*(c + b*x)]))/(12*b)
Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {5091, 3042, 3086, 15, 4255, 3042, 4257}
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.
\(\displaystyle \int \sin (a+b x) \sec ^4(b x+c) \, dx\) |
\(\Big \downarrow \) 5091 |
\(\displaystyle \sin (a-c) \int \sec ^3(c+b x)dx+\cos (a-c) \int \sec ^3(c+b x) \tan (c+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx+\cos (a-c) \int \sec (c+b x)^3 \tan (c+b x)dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\cos (a-c) \int \sec ^2(c+b x)d\sec (c+b x)}{b}+\sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )^3dx+\frac {\cos (a-c) \sec ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \sin (a-c) \left (\frac {1}{2} \int \sec (c+b x)dx+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )+\frac {\cos (a-c) \sec ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin (a-c) \left (\frac {1}{2} \int \csc \left (c+b x+\frac {\pi }{2}\right )dx+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )+\frac {\cos (a-c) \sec ^3(b x+c)}{3 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \sin (a-c) \left (\frac {\text {arctanh}(\sin (b x+c))}{2 b}+\frac {\tan (b x+c) \sec (b x+c)}{2 b}\right )+\frac {\cos (a-c) \sec ^3(b x+c)}{3 b}\) |
Input:
Int[Sec[c + b*x]^4*Sin[a + b*x],x]
Output:
(Cos[a - c]*Sec[c + b*x]^3)/(3*b) + Sin[a - c]*(ArcTanh[Sin[c + b*x]]/(2*b ) + (Sec[c + b*x]*Tan[c + b*x])/(2*b))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[Sec[w_]^(n_.)*Sin[v_], x_Symbol] :> Simp[Cos[v - w] Int[Tan[w]*Sec[w] ^(n - 1), x], x] + Simp[Sin[v - w] Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0 ] && FreeQ[v - w, x] && NeQ[w, v]
Result contains complex when optimal does not.
Time = 10.04 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.85
method | result | size |
risch | \(\frac {-3 \,{\mathrm e}^{i \left (5 b x +7 a +4 c \right )}+3 \,{\mathrm e}^{i \left (5 b x +5 a +6 c \right )}+8 \,{\mathrm e}^{i \left (3 b x +7 a +2 c \right )}+8 \,{\mathrm e}^{i \left (3 b x +5 a +4 c \right )}+3 \,{\mathrm e}^{i \left (b x +7 a \right )}-3 \,{\mathrm e}^{i \left (b x +5 a +2 c \right )}}{6 b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}\) | \(191\) |
default | \(\text {Expression too large to display}\) | \(1782\) |
Input:
int(sec(b*x+c)^4*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/6/b/(exp(2*I*(b*x+a+c))+exp(2*I*a))^3*(-3*exp(I*(5*b*x+7*a+4*c))+3*exp(I *(5*b*x+5*a+6*c))+8*exp(I*(3*b*x+7*a+2*c))+8*exp(I*(3*b*x+5*a+4*c))+3*exp( I*(b*x+7*a))-3*exp(I*(b*x+5*a+2*c)))+1/2*ln(exp(I*(b*x+a))+I*exp(I*(a-c))) /b*sin(a-c)-1/2*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))/b*sin(a-c)
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=-\frac {3 \, \cos \left (b x + c\right )^{3} \log \left (\sin \left (b x + c\right ) + 1\right ) \sin \left (-a + c\right ) - 3 \, \cos \left (b x + c\right )^{3} \log \left (-\sin \left (b x + c\right ) + 1\right ) \sin \left (-a + c\right ) + 6 \, \cos \left (b x + c\right ) \sin \left (b x + c\right ) \sin \left (-a + c\right ) - 4 \, \cos \left (-a + c\right )}{12 \, b \cos \left (b x + c\right )^{3}} \] Input:
integrate(sec(b*x+c)^4*sin(b*x+a),x, algorithm="fricas")
Output:
-1/12*(3*cos(b*x + c)^3*log(sin(b*x + c) + 1)*sin(-a + c) - 3*cos(b*x + c) ^3*log(-sin(b*x + c) + 1)*sin(-a + c) + 6*cos(b*x + c)*sin(b*x + c)*sin(-a + c) - 4*cos(-a + c))/(b*cos(b*x + c)^3)
Timed out. \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\text {Timed out} \] Input:
integrate(sec(b*x+c)**4*sin(b*x+a),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1424 vs. \(2 (61) = 122\).
Time = 0.21 (sec) , antiderivative size = 1424, normalized size of antiderivative = 21.25 \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:
integrate(sec(b*x+c)^4*sin(b*x+a),x, algorithm="maxima")
Output:
-1/12*(2*(3*cos(5*b*x + 2*a + 4*c) - 3*cos(5*b*x + 6*c) - 8*cos(3*b*x + 2* a + 2*c) - 8*cos(3*b*x + 4*c) - 3*cos(b*x + 2*a) + 3*cos(b*x + 2*c))*cos(6 *b*x + a + 6*c) + 6*(3*cos(4*b*x + a + 4*c) + 3*cos(2*b*x + a + 2*c) + cos (a))*cos(5*b*x + 2*a + 4*c) - 6*(3*cos(4*b*x + a + 4*c) + 3*cos(2*b*x + a + 2*c) + cos(a))*cos(5*b*x + 6*c) - 6*(8*cos(3*b*x + 2*a + 2*c) + 8*cos(3* b*x + 4*c) + 3*cos(b*x + 2*a) - 3*cos(b*x + 2*c))*cos(4*b*x + a + 4*c) - 1 6*(3*cos(2*b*x + a + 2*c) + cos(a))*cos(3*b*x + 2*a + 2*c) - 16*(3*cos(2*b *x + a + 2*c) + cos(a))*cos(3*b*x + 4*c) - 18*(cos(b*x + 2*a) - cos(b*x + 2*c))*cos(2*b*x + a + 2*c) - 6*cos(b*x + 2*a)*cos(a) + 6*cos(b*x + 2*c)*co s(a) - 3*(cos(6*b*x + a + 6*c)^2*sin(-a + c) + 9*cos(4*b*x + a + 4*c)^2*si n(-a + c) + 9*cos(2*b*x + a + 2*c)^2*sin(-a + c) + 6*cos(2*b*x + a + 2*c)* cos(a)*sin(-a + c) + sin(6*b*x + a + 6*c)^2*sin(-a + c) + 9*sin(4*b*x + a + 4*c)^2*sin(-a + c) + 9*sin(2*b*x + a + 2*c)^2*sin(-a + c) + 6*sin(2*b*x + a + 2*c)*sin(a)*sin(-a + c) + 2*(3*cos(4*b*x + a + 4*c)*sin(-a + c) + 3* cos(2*b*x + a + 2*c)*sin(-a + c) + cos(a)*sin(-a + c))*cos(6*b*x + a + 6*c ) + 6*(3*cos(2*b*x + a + 2*c)*sin(-a + c) + cos(a)*sin(-a + c))*cos(4*b*x + a + 4*c) + 2*(3*sin(4*b*x + a + 4*c)*sin(-a + c) + 3*sin(2*b*x + a + 2*c )*sin(-a + c) + sin(a)*sin(-a + c))*sin(6*b*x + a + 6*c) + 6*(3*sin(2*b*x + a + 2*c)*sin(-a + c) + sin(a)*sin(-a + c))*sin(4*b*x + a + 4*c) + (cos(a )^2 + sin(a)^2)*sin(-a + c))*log((cos(b*x + 2*c)^2 + cos(c)^2 - 2*cos(c...
Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (61) = 122\).
Time = 0.14 (sec) , antiderivative size = 495, normalized size of antiderivative = 7.39 \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:
integrate(sec(b*x+c)^4*sin(b*x+a),x, algorithm="giac")
Output:
1/3*(3*(tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - t an(1/2*c))*log(abs(tan(1/2*b*x + 1/2*c) + 1))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) - 3*(tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a )*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*log(abs(tan(1/2*b*x + 1/2*c) - 1 ))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) + 2*(3*ta n(1/2*b*x + 1/2*c)^5*tan(1/2*a)^2*tan(1/2*c) - 3*tan(1/2*b*x + 1/2*c)^5*ta n(1/2*a)*tan(1/2*c)^2 - 3*tan(1/2*b*x + 1/2*c)^4*tan(1/2*a)^2*tan(1/2*c)^2 + 3*tan(1/2*b*x + 1/2*c)^5*tan(1/2*a) + 3*tan(1/2*b*x + 1/2*c)^4*tan(1/2* a)^2 - 3*tan(1/2*b*x + 1/2*c)^5*tan(1/2*c) - 12*tan(1/2*b*x + 1/2*c)^4*tan (1/2*a)*tan(1/2*c) + 3*tan(1/2*b*x + 1/2*c)^4*tan(1/2*c)^2 - 3*tan(1/2*b*x + 1/2*c)^4 - 3*tan(1/2*b*x + 1/2*c)*tan(1/2*a)^2*tan(1/2*c) + 3*tan(1/2*b *x + 1/2*c)*tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*a)^2*tan(1/2*c)^2 - 3*tan(1/ 2*b*x + 1/2*c)*tan(1/2*a) + tan(1/2*a)^2 + 3*tan(1/2*b*x + 1/2*c)*tan(1/2* c) - 4*tan(1/2*a)*tan(1/2*c) + tan(1/2*c)^2 - 1)/((tan(1/2*a)^2*tan(1/2*c) ^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*(tan(1/2*b*x + 1/2*c)^2 - 1)^3))/b
Timed out. \[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\text {Hanged} \] Input:
int(sin(a + b*x)/cos(c + b*x)^4,x)
Output:
\text{Hanged}
\[ \int \sec ^4(c+b x) \sin (a+b x) \, dx=\int \sec \left (b x +c \right )^{4} \sin \left (b x +a \right )d x \] Input:
int(sec(b*x+c)^4*sin(b*x+a),x)
Output:
int(sec(b*x + c)**4*sin(a + b*x),x)