\(\int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx\) [12]

Optimal result

Integrand size = 19, antiderivative size = 84 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))} \]

[Out]

3/8*a*arctanh(sin(d*x+c))/d+1/8*a^3/d/(a-a*sin(d*x+c))^2+1/4*a^2/d/(a-a*sin(d*x+c))-1/8*a^2/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2746, 46, 212} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}+\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d} \]

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + a^2/(4*d*(a - a*Sin[c + d*x])) - a^2/(8
*d*(a + a*Sin[c + d*x]))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {a}{4 \cos \left (d x +c \right )^{4}}+a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(63\)
default	\(\frac {\frac {a}{4 \cos \left (d x +c \right )^{4}}+a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(63\)
risch	\(-\frac {i a \left (-6 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+6 i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4} d}-\frac {3 a \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(133\)
parallelrisch	\(\frac {3 \left (\left (-\cos \left (2 d x +2 c \right )+\frac {\sin \left (d x +c \right )}{2}+\frac {\sin \left (3 d x +3 c \right )}{2}-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {\cos \left (2 d x +2 c \right )}{3}+\frac {7 \sin \left (d x +c \right )}{3}+\frac {\sin \left (3 d x +3 c \right )}{3}-\frac {1}{3}\right ) a}{4 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\)	\(161\)
norman	\(\frac {\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(221\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.62 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{2} - 3 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a \sin \left (d x + c\right ) - 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 2 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.10 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {6 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a \sin \left (d x + c\right ) + 5 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {9 \, a \sin \left (d x + c\right )^{2} - 26 \, a \sin \left (d x + c\right ) + 21 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

[In]

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

[Out]

1/32*(6*a*log(abs(sin(d*x + c) + 1)) - 6*a*log(abs(sin(d*x + c) - 1)) - 2*(3*a*sin(d*x + c) + 5*a)/(sin(d*x +
c) + 1) + (9*a*sin(d*x + c)^2 - 26*a*sin(d*x + c) + 21*a)/(sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 6.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3\,a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{8\,d}-\frac {-\frac {3\,a\,{\sin \left (c+d\,x\right )}^2}{8}+\frac {3\,a\,\sin \left (c+d\,x\right )}{8}+\frac {a}{4}}{d\,\left (-{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \]

[In]

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

[Out]

(3*a*atanh(sin(c + d*x)))/(8*d) - (a/4 + (3*a*sin(c + d*x))/8 - (3*a*sin(c + d*x)^2)/8)/(d*(sin(c + d*x) + sin
(c + d*x)^2 - sin(c + d*x)^3 - 1))