\(\int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx\) [1484]

Optimal result

Integrand size = 25, antiderivative size = 103 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \log (\cos (c+d x))}{d}-\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d} \]

[Out]

3/8*a*arctanh(sin(d*x+c))/d-b*ln(cos(d*x+c))/d-3/8*a*sec(d*x+c)*tan(d*x+c)/d-1/2*b*tan(d*x+c)^2/d+1/4*a*sec(d*
x+c)*tan(d*x+c)^3/d+1/4*b*tan(d*x+c)^4/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2913, 2691, 3855, 3554, 3556} \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}-\frac {b \log (\cos (c+d x))}{d} \]

[In]

Int[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Log[Cos[c + d*x]])/d - (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (b*Tan[c
+ d*x]^2)/(2*d) + (a*Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) + (b*Tan[c + d*x]^4)/(4*d)

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2913

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = a \int \sec (c+d x) \tan ^4(c+d x) \, dx+b \int \tan ^5(c+d x) \, dx \\ & = \frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {1}{4} (3 a) \int \sec (c+d x) \tan ^2(c+d x) \, dx-b \int \tan ^3(c+d x) \, dx \\ & = -\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d}+\frac {1}{8} (3 a) \int \sec (c+d x) \, dx+b \int \tan (c+d x) \, dx \\ & = \frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \log (\cos (c+d x))}{d}-\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.12 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {b \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]

[In]

Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (3*a*Sec[c + d*x]^3*Tan[c + d*x])/
(4*d) + (a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (b*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d
)

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(111\)
default	\(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(111\)
parallelrisch	\(\frac {16 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (a +\frac {8 b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {8 b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 a \sin \left (d x +c \right )-5 a \sin \left (3 d x +3 c \right )-4 b \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right ) b +b}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(196\)
risch	\(i x b +\frac {2 i b c}{d}+\frac {i \left (5 a \,{\mathrm e}^{7 i \left (d x +c \right )}-3 a \,{\mathrm e}^{5 i \left (d x +c \right )}+16 i b \,{\mathrm e}^{6 i \left (d x +c \right )}+3 a \,{\mathrm e}^{3 i \left (d x +c \right )}+16 i b \,{\mathrm e}^{4 i \left (d x +c \right )}-5 a \,{\mathrm e}^{i \left (d x +c \right )}+16 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}\)	\(198\)
norman	\(\frac {-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {6 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (3 a -8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (3 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\)	\(252\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec
(d*x+c)+tan(d*x+c)))+b*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 16 \, b \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*((3*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*a + 8*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 1
6*b*cos(d*x + c)^2 - 2*(5*a*cos(d*x + c)^2 - 2*a)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**4*(a+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, a \sin \left (d x + c\right )^{3} + 8 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 6 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*((3*a - 8*b)*log(sin(d*x + c) + 1) - (3*a + 8*b)*log(sin(d*x + c) - 1) + 2*(5*a*sin(d*x + c)^3 + 8*b*sin(
d*x + c)^2 - 3*a*sin(d*x + c) - 6*b)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (6 \, b \sin \left (d x + c\right )^{4} + 5 \, a \sin \left (d x + c\right )^{3} - 4 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*((3*a - 8*b)*log(abs(sin(d*x + c) + 1)) - (3*a + 8*b)*log(abs(sin(d*x + c) - 1)) + 2*(6*b*sin(d*x + c)^4
+ 5*a*sin(d*x + c)^3 - 4*b*sin(d*x + c)^2 - 3*a*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 12.12 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a}{8}+b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a}{8}-b\right )}{d} \]

[In]

int((sin(c + d*x)^4*(a + b*sin(c + d*x)))/cos(c + d*x)^5,x)

[Out]

(b*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ((3*a*tan(c/2 + (d*x)/2))/4 - (11*a*tan(c/2 + (d*x)/2)^3)/4 - (11*a*tan(
c/2 + (d*x)/2)^5)/4 + (3*a*tan(c/2 + (d*x)/2)^7)/4 + 2*b*tan(c/2 + (d*x)/2)^2 - 8*b*tan(c/2 + (d*x)/2)^4 + 2*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)) - (log(tan(c/2 + (d*x)/2) - 1)*((3*a)/8 + b))/d + (log(tan(c/2 + (d*x)/2) + 1)*((3*a)/8 - b
))/d