\(\int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\) [23]

Optimal result

Integrand size = 21, antiderivative size = 64 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {a^3}{4 d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2746, 46, 212} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {a^3}{4 d (a-a \sin (c+d x))}+\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d} \]

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*ArcTanh[Sin[c + d*x]])/(4*d) + a^4/(4*d*(a - a*Sin[c + d*x])^2) + a^3/(4*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)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {1}{2 a (a-x)^3}+\frac {1}{4 a^2 (a-x)^2}+\frac {1}{4 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {a^3}{4 d (a-a \sin (c+d x))}+\frac {a^3 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 d} \\ & = \frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {a^3}{4 d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sec ^4(c+d x) \left (2+\text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x))^2-\sin (c+d x)\right ) (1+\sin (c+d x))^2}{4 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Sec[c + d*x]^4*(2 + ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])^2 - Sin[c + d*x])*(1 + Sin[c + d*x])^2)/(4*
d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.56

method	result	size
risch	\(-\frac {i a^{2} \left (-4 i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4}}-\frac {a^{2} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\)	\(100\)
parallelrisch	\(-\frac {\left (\left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \cos \left (2 d x +2 c \right )+6 \sin \left (d x +c \right )-2\right ) a^{2}}{4 d \left (-3+\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right )}\)	\(117\)
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2}}{2 \cos \left (d x +c \right )^{4}}+a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(132\)
default	\(\frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2}}{2 \cos \left (d x +c \right )^{4}}+a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(132\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {11 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {9 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {9 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {3 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(281\)

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*I*a^2/d/(-I+exp(I*(d*x+c)))^4*(-4*I*exp(2*I*(d*x+c))+exp(3*I*(d*x+c))-exp(I*(d*x+c)))-1/4*a^2/d*ln(-I+exp
(I*(d*x+c)))+1/4*a^2/d*ln(exp(I*(d*x+c))+I)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (60) = 120\).

Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.95 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int 2 \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*sec(c + d*x)**5, x) + Integral(sin(c + d*x)**2*sec(c + d*x)**5, x) + Integral(se
c(c + d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*(a^2*log(sin(d*x + c) + 1) - a^2*log(sin(d*x + c) - 1) - 2*(a^2*sin(d*x + c) - 2*a^2)/(sin(d*x + c)^2 - 2*
sin(d*x + c) + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.20 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} - 10 \, a^{2} \sin \left (d x + c\right ) + 11 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) - 2*a^2*log(abs(sin(d*x + c) - 1)) + (3*a^2*sin(d*x + c)^2 - 10*a^2*sin
(d*x + c) + 11*a^2)/(sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,d}-\frac {\frac {a^2\,\sin \left (c+d\,x\right )}{4}-\frac {a^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )} \]

[In]

int((a + a*sin(c + d*x))^2/cos(c + d*x)^5,x)

[Out]

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