\(\int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx\) [39]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^6}{6 d (a-a \sin (c+d x))^3}+\frac {a^5}{8 d (a-a \sin (c+d x))^2}+\frac {a^4}{8 d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, 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} \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^6}{6 d (a-a \sin (c+d x))^3}+\frac {a^5}{8 d (a-a \sin (c+d x))^2}+\frac {a^4}{8 d (a-a \sin (c+d x))}+\frac {a^3 \text {arctanh}(\sin (c+d x))}{8 d} \]

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*ArcTanh[Sin[c + d*x]])/(8*d) + a^6/(6*d*(a - a*Sin[c + d*x])^3) + a^5/(8*d*(a - a*Sin[c + d*x])^2) + a^4/
(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^7 \text {Subst}\left (\int \frac {1}{(a-x)^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {1}{2 a (a-x)^4}+\frac {1}{4 a^2 (a-x)^3}+\frac {1}{8 a^3 (a-x)^2}+\frac {1}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6}{6 d (a-a \sin (c+d x))^3}+\frac {a^5}{8 d (a-a \sin (c+d x))^2}+\frac {a^4}{8 d (a-a \sin (c+d x))}+\frac {a^4 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d} \\ & = \frac {a^3 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^6}{6 d (a-a \sin (c+d x))^3}+\frac {a^5}{8 d (a-a \sin (c+d x))^2}+\frac {a^4}{8 d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \sec ^6(c+d x) (1+\sin (c+d x))^3 \left (-10+3 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x))^3+9 \sin (c+d x)-3 \sin ^2(c+d x)\right )}{24 d} \]

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3,x]

[Out]

-1/24*(a^3*Sec[c + d*x]^6*(1 + Sin[c + d*x])^3*(-10 + 3*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])^3 + 9*Sin[c
+ d*x] - 3*Sin[c + d*x]^2))/d

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.44

method	result	size
risch	\(-\frac {i a^{3} \left (-18 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+18 i {\mathrm e}^{2 i \left (d x +c \right )}-46 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{12 d \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{6}}-\frac {a^{3} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(125\)
parallelrisch	\(\frac {3 \left (\left (-\cos \left (2 d x +2 c \right )-\frac {5 \sin \left (d x +c \right )}{2}+\frac {\sin \left (3 d x +3 c \right )}{6}+\frac {5}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )+\frac {5 \sin \left (d x +c \right )}{2}-\frac {\sin \left (3 d x +3 c \right )}{6}-\frac {5}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \cos \left (2 d x +2 c \right )-\frac {19 \sin \left (d x +c \right )}{3}+\frac {5 \sin \left (3 d x +3 c \right )}{9}+3\right ) a^{3}}{4 d \left (-10+15 \sin \left (d x +c \right )-\sin \left (3 d x +3 c \right )+6 \cos \left (2 d x +2 c \right )\right )}\)	\(163\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a^{3}}{2 \cos \left (d x +c \right )^{6}}+a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(202\)
default	\(\frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a^{3}}{2 \cos \left (d x +c \right )^{6}}+a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(202\)

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-1/12*I*a^3/d/(-I+exp(I*(d*x+c)))^6*(-18*I*exp(4*I*(d*x+c))+3*exp(5*I*(d*x+c))+18*I*exp(2*I*(d*x+c))-46*exp(3*
I*(d*x+c))+3*exp(I*(d*x+c)))-1/8*a^3/d*ln(-I+exp(I*(d*x+c)))+1/8*a^3/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. 185 vs. \(2 (82) = 164\).

Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.13 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \cos \left (d x + c\right )^{2} + 18 \, a^{3} \sin \left (d x + c\right ) - 26 \, a^{3} + 3 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{48 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/48*(6*a^3*cos(d*x + c)^2 + 18*a^3*sin(d*x + c) - 26*a^3 + 3*(3*a^3*cos(d*x + c)^2 - 4*a^3 - (a^3*cos(d*x + c
)^2 - 4*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) - 3*(3*a^3*cos(d*x + c)^2 - 4*a^3 - (a^3*cos(d*x + c)^2 - 4*a
^3)*sin(d*x + c))*log(-sin(d*x + c) + 1))/(3*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3 \, a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/48*(3*a^3*log(sin(d*x + c) + 1) - 3*a^3*log(sin(d*x + c) - 1) - 2*(3*a^3*sin(d*x + c)^2 - 9*a^3*sin(d*x + c)
+ 10*a^3)/(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c) - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {11 \, a^{3} \sin \left (d x + c\right )^{3} - 45 \, a^{3} \sin \left (d x + c\right )^{2} + 69 \, a^{3} \sin \left (d x + c\right ) - 51 \, a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(6*a^3*log(abs(sin(d*x + c) + 1)) - 6*a^3*log(abs(sin(d*x + c) - 1)) + (11*a^3*sin(d*x + c)^3 - 45*a^3*si
n(d*x + c)^2 + 69*a^3*sin(d*x + c) - 51*a^3)/(sin(d*x + c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{8\,d}-\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{8}-\frac {3\,a^3\,\sin \left (c+d\,x\right )}{8}+\frac {5\,a^3}{12}}{d\,\left ({\sin \left (c+d\,x\right )}^3-3\,{\sin \left (c+d\,x\right )}^2+3\,\sin \left (c+d\,x\right )-1\right )} \]

[In]

int((a + a*sin(c + d*x))^3/cos(c + d*x)^7,x)

[Out]

(a^3*atanh(sin(c + d*x)))/(8*d) - ((5*a^3)/12 - (3*a^3*sin(c + d*x))/8 + (a^3*sin(c + d*x)^2)/8)/(d*(3*sin(c +
d*x) - 3*sin(c + d*x)^2 + sin(c + d*x)^3 - 1))