\(\int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx\) [871]

Optimal result

Integrand size = 27, antiderivative size = 96 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {6 a^3 \log (1-\sin (c+d x))}{d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {4 a^4}{d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 45} \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {4 a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \sin ^2(c+d x)}{2 d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {6 a^3 \log (1-\sin (c+d x))}{d} \]

[In]

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

[Out]

(-6*a^3*Log[1 - Sin[c + d*x]])/d - (3*a^3*Sin[c + d*x])/d - (a^3*Sin[c + d*x]^2)/(2*d) + a^5/(2*d*(a - a*Sin[c
+ d*x])^2) - (4*a^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 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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^4}{a^4 (a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \frac {x^4}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (-3 a+\frac {a^4}{(a-x)^3}-\frac {4 a^3}{(a-x)^2}+\frac {6 a^2}{a-x}-x\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {6 a^3 \log (1-\sin (c+d x))}{d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {4 a^4}{d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {a^3 \left (12 \log (1-\sin (c+d x))+\frac {7-8 \sin (c+d x)}{(-1+\sin (c+d x))^2}+6 \sin (c+d x)+\sin ^2(c+d x)\right )}{2 d} \]

[In]

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

-1/2*(a^3*(12*Log[1 - Sin[c + d*x]] + (7 - 8*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 6*Sin[c + d*x] + Sin[c + d*
x]^2))/d

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.45

method	result	size
parallelrisch	\(\frac {6 \left (\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 \cos \left (2 d x +2 c \right )+6-8 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {17 \cos \left (2 d x +2 c \right )}{12}+\frac {\cos \left (4 d x +4 c \right )}{48}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {\sin \left (3 d x +3 c \right )}{6}-\frac {23}{16}\right ) a^{3}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(139\)
risch	\(6 i a^{3} x +\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {12 i a^{3} c}{d}+\frac {2 i a^{3} \left (-7 i {\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}-\frac {12 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(168\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(282\)
default	\(\frac {a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(282\)
norman	\(\frac {\frac {64 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {64 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {44 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {48 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {44 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {12 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {6 a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(321\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

6*((cos(2*d*x+2*c)-3+4*sin(d*x+c))*ln(sec(1/2*d*x+1/2*c)^2)+(-2*cos(2*d*x+2*c)+6-8*sin(d*x+c))*ln(tan(1/2*d*x+
1/2*c)-1)+17/12*cos(2*d*x+2*c)+1/48*cos(4*d*x+4*c)+5/2*sin(d*x+c)-1/6*sin(3*d*x+3*c)-23/16)*a^3/d/(cos(2*d*x+2
*c)-3+4*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.33 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {2 \, a^{3} \cos \left (d x + c\right )^{4} + 19 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3} - 24 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, a^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(2*a^3*cos(d*x + c)^4 + 19*a^3*cos(d*x + c)^2 - 8*a^3 - 24*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^
3)*log(-sin(d*x + c) + 1) - 2*(4*a^3*cos(d*x + c)^2 - 3*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c
) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=-\frac {a^{3} \sin \left (d x + c\right )^{2} + 12 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {8 \, a^{3} \sin \left (d x + c\right ) - 7 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(a^3*sin(d*x + c)^2 + 12*a^3*log(sin(d*x + c) - 1) + 6*a^3*sin(d*x + c) - (8*a^3*sin(d*x + c) - 7*a^3)/(s
in(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (94) = 188\).

Time = 0.51 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.18 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {6 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac {25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 106 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 164 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 106 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(6*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (9*a^3*tan(1/2*d*x + 1/2*
c)^4 + 6*a^3*tan(1/2*d*x + 1/2*c)^3 + 20*a^3*tan(1/2*d*x + 1/2*c)^2 + 6*a^3*tan(1/2*d*x + 1/2*c) + 9*a^3)/(tan
(1/2*d*x + 1/2*c)^2 + 1)^2 + (25*a^3*tan(1/2*d*x + 1/2*c)^4 - 106*a^3*tan(1/2*d*x + 1/2*c)^3 + 164*a^3*tan(1/2
*d*x + 1/2*c)^2 - 106*a^3*tan(1/2*d*x + 1/2*c) + 25*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 12.68 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.74 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx=\frac {6\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {12\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-36\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+52\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-64\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+52\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-36\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+12\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {12\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d} \]

[In]

int((sin(c + d*x)^4*(a + a*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(6*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (52*a^3*tan(c/2 + (d*x)/2)^3 - 36*a^3*tan(c/2 + (d*x)/2)^2 - 64*a^3*
tan(c/2 + (d*x)/2)^4 + 52*a^3*tan(c/2 + (d*x)/2)^5 - 36*a^3*tan(c/2 + (d*x)/2)^6 + 12*a^3*tan(c/2 + (d*x)/2)^7
+ 12*a^3*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 12*tan(c/2 + (d*x)/2)^3 + 14
*tan(c/2 + (d*x)/2)^4 - 12*tan(c/2 + (d*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^6 - 4*tan(c/2 + (d*x)/2)^7 + tan(c/2 +
(d*x)/2)^8 + 1)) - (12*a^3*log(tan(c/2 + (d*x)/2) - 1))/d