\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx\) [863]

Optimal result

Integrand size = 29, antiderivative size = 87 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {5 a^3}{4 d (a-a \sin (c+d x))} \]

[Out]

-7/8*a^2*ln(1-sin(d*x+c))/d-1/8*a^2*ln(1+sin(d*x+c))/d+1/4*a^4/d/(a-a*sin(d*x+c))^2-5/4*a^3/d/(a-a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 90} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {5 a^3}{4 d (a-a \sin (c+d x))}-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2 \left (3 \text {arctanh}(\sin (c+d x))-6 \sec ^3(c+d x) \tan (c+d x)+\sec (c+d x) \tan (c+d x) \left (3+8 \tan ^2(c+d x)\right )-2 \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)-\tan ^4(c+d x)\right )\right )}{4 d} \]

[In]

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

[Out]

(a^2*(3*ArcTanh[Sin[c + d*x]] - 6*Sec[c + d*x]^3*Tan[c + d*x] + Sec[c + d*x]*Tan[c + d*x]*(3 + 8*Tan[c + d*x]^
2) - 2*(2*Log[Cos[c + d*x]] + Tan[c + d*x]^2 - Tan[c + d*x]^4)))/(4*d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.44

method	result	size
risch	\(i a^{2} x +\frac {2 i a^{2} c}{d}+\frac {i \left (-8 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2} {\mathrm e}^{i \left (d x +c \right )}+5 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\)	\(125\)
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(137\)
default	\(\frac {a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+2 a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(137\)
parallelrisch	\(-\frac {7 \left (\frac {4}{7}+\frac {4 \left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{7}-\frac {4 \cos \left (2 d x +2 c \right )}{7}-\frac {6 \sin \left (d x +c \right )}{7}\right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(151\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {5 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {15 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {2 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(302\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

I*a^2*x+2*I*a^2/d*c+1/2*I*(-8*I*a^2*exp(2*I*(d*x+c))-5*a^2*exp(I*(d*x+c))+5*a^2*exp(3*I*(d*x+c)))/(exp(I*(d*x+
c))-I)^4/d-1/4*a^2/d*ln(exp(I*(d*x+c))+I)-7/4*a^2/d*ln(exp(I*(d*x+c))-I)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.44 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {10 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/8*(10*a^2*sin(d*x + c) - 8*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) +
7*(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. \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8*(a^2*log(sin(d*x + c) + 1) + 7*a^2*log(sin(d*x + c) - 1) - 2*(5*a^2*sin(d*x + c) - 4*a^2)/(sin(d*x + c)^2
- 2*sin(d*x + c) + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 14 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {21 \, a^{2} \sin \left (d x + c\right )^{2} - 22 \, a^{2} \sin \left (d x + c\right ) + 5 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) + 14*a^2*log(abs(sin(d*x + c) - 1)) - (21*a^2*sin(d*x + c)^2 - 22*a^2*
sin(d*x + c) + 5*a^2)/(sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 10.50 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.91 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {7\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{4\,d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{4\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \]

[In]

int((sin(c + d*x)^3*(a + a*sin(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (7*a^2*log(tan(c/2 + (d*x)/2) - 1))/(4*d) - (a^2*log(tan(c/2 + (d*x)/2
) + 1))/(4*d) - ((3*a^2*tan(c/2 + (d*x)/2)^3)/2 - 4*a^2*tan(c/2 + (d*x)/2)^2 + (3*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))