\(\int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx\) [1487]

Optimal result

Integrand size = 25, antiderivative size = 74 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2913, 2686, 30, 2691, 3853, 3855} \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=\frac {a \sec ^4(c+d x)}{4 d}-\frac {b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

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

[Out]

-1/8*(b*ArcTanh[Sin[c + d*x]])/d + (a*Sec[c + d*x]^4)/(4*d) - (b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c +
d*x]^3*Tan[c + d*x])/(4*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[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])

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 3853

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] + Dist[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]

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 ^4(c+d x) \tan (c+d x) \, dx+b \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx \\ & = \frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{4} b \int \sec ^3(c+d x) \, dx+\frac {a \text {Subst}\left (\int x^3 \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {a \sec ^4(c+d x)}{4 d}-\frac {b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{8} b \int \sec (c+d x) \, dx \\ & = -\frac {b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[In]

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

[Out]

-1/8*(b*ArcTanh[Sin[c + d*x]])/d + (a*Sec[c + d*x]^4)/(4*d) - (b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c +
d*x]^3*Tan[c + d*x])/(4*d)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {a}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(80\)
default	\(\frac {\frac {a}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(80\)
risch	\(\frac {i \left (b \,{\mathrm e}^{7 i \left (d x +c \right )}-16 i a \,{\mathrm e}^{4 i \left (d x +c \right )}-7 b \,{\mathrm e}^{5 i \left (d x +c \right )}+7 b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}\)	\(120\)
parallelrisch	\(\frac {2 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 a \cos \left (2 d x +2 c \right )-\cos \left (4 d x +4 c \right ) a +7 b \sin \left (d x +c \right )-b \sin \left (3 d x +3 c \right )+5 a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(153\)
norman	\(\frac {\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(221\)

[In]

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

[Out]

1/d*(1/4*a/cos(d*x+c)^4+b*(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(s
ec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \log \left (\sin \left (d x + c\right ) + 1\right ) - b \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (b \sin \left (d x + c\right )^{3} + b \sin \left (d x + c\right ) + 2 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - b \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (b \sin \left (d x + c\right )^{3} + b \sin \left (d x + c\right ) + 2 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.13 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.14 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]

[In]

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

[Out]

((b*tan(c/2 + (d*x)/2))/4 + 2*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)^6 + (7*b*tan(c/2 + (d*x)/2)^3)/4
+ (7*b*tan(c/2 + (d*x)/2)^5)/4 + (b*tan(c/2 + (d*x)/2)^7)/4)/(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)) - (b*atanh(tan(c/2 + (d*x)/2)))/(4*d)