\(\int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx\) [864]

Optimal result

Integrand size = 29, antiderivative size = 64 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {3 a^3}{4 d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2915, 12, 90, 212} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {3 a^3}{4 d (a-a \sin (c+d x))}+\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d} \]

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2*Tan[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) - (3*a^3)/(4*d*(a - a*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 2915

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.61 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \left (\text {arctanh}(\sin (c+d x))+\frac {-2+3 \sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{4 d} \]

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]

[Out]

(a^2*(ArcTanh[Sin[c + d*x]] + (-2 + 3*Sin[c + d*x])/(-1 + Sin[c + d*x])^2))/(4*d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.59

method	result	size
risch	\(\frac {i a^{2} \left (-4 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\)	\(102\)
parallelrisch	\(-\frac {\left (\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \cos \left (2 d x +2 c \right )-2 \sin \left (d x +c \right )+2\right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(117\)
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}+a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(167\)
default	\(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}+a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(167\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {7 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {13 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {16 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(243\)

[In]

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

[Out]

1/2*I*a^2/d/(exp(I*(d*x+c))-I)^4*(-4*I*exp(2*I*(d*x+c))+3*exp(3*I*(d*x+c))-3*exp(I*(d*x+c)))+1/4*a^2/d*ln(exp(
I*(d*x+c))+I)-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 = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.95 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {6 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

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

[Out]

-1/8*(6*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(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(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*(3*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 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right ) - 5 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(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 + 6*a^2*sin(
d*x + c) - 5*a^2)/(sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 12.00 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

[In]

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

[Out]

(a^2*atanh(tan(c/2 + (d*x)/2)))/(2*d) - ((a^2*tan(c/2 + (d*x)/2)^3)/2 - 2*a^2*tan(c/2 + (d*x)/2)^2 + (a^2*tan(
c/2 + (d*x)/2))/2)/(d*(6*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x
)/2)^4 + 1))