3.50 \(\int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx\)

Optimal. Leaf size=138 \[ -\frac {a^3 \sin ^3(c+d x)}{d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {33 a^3 x}{8} \]

[Out]

-33/8*a^3*x+3/2*a^3*arctanh(sin(d*x+c))/d-2*a^3*sin(d*x+c)/d+7/8*a^3*cos(d*x+c)*sin(d*x+c)/d+1/4*a^3*cos(d*x+c
)^3*sin(d*x+c)/d-a^3*sin(d*x+c)^3/d+3*a^3*tan(d*x+c)/d+1/2*a^3*sec(d*x+c)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3872, 2872, 2637, 2635, 8, 2633, 3770, 3767, 3768} \[ -\frac {a^3 \sin ^3(c+d x)}{d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {33 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-33*a^3*x)/8 + (3*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (2*a^3*Sin[c + d*x])/d + (7*a^3*Cos[c + d*x]*Sin[c + d*x
])/(8*d) + (a^3*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a^3*Sin[c + d*x]^3)/d + (3*a^3*Tan[c + d*x])/d + (a^3*Se
c[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2872

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^3 \sin ^4(c+d x) \, dx &=-\int (-a-a \cos (c+d x))^3 \sin (c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac {\int \left (5 a^7+5 a^7 \cos (c+d x)-a^7 \cos ^2(c+d x)-3 a^7 \cos ^3(c+d x)-a^7 \cos ^4(c+d x)-a^7 \sec (c+d x)-3 a^7 \sec ^2(c+d x)-a^7 \sec ^3(c+d x)\right ) \, dx}{a^4}\\ &=-5 a^3 x+a^3 \int \cos ^2(c+d x) \, dx+a^3 \int \cos ^4(c+d x) \, dx+a^3 \int \sec (c+d x) \, dx+a^3 \int \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx-\left (5 a^3\right ) \int \cos (c+d x) \, dx\\ &=-5 a^3 x+\frac {a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 a^3 \sin (c+d x)}{d}+\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^3 \int 1 \, dx+\frac {1}{2} a^3 \int \sec (c+d x) \, dx+\frac {1}{4} \left (3 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (3 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-\frac {9 a^3 x}{2}+\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{8} \left (3 a^3\right ) \int 1 \, dx\\ &=-\frac {33 a^3 x}{8}+\frac {3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {2 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \sin ^3(c+d x)}{d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.41, size = 114, normalized size = 0.83 \[ \frac {a^3 \sec ^2(c+d x) \left (-16 \sin (c+d x)+225 \sin (2 (c+d x))-72 \sin (3 (c+d x))+18 \sin (4 (c+d x))+8 \sin (5 (c+d x))+\sin (6 (c+d x))-264 (c+d x) \cos (2 (c+d x))+192 \cos ^2(c+d x) \tanh ^{-1}(\sin (c+d x))-264 c-264 d x\right )}{128 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c + d*x]^2*(-264*c - 264*d*x + 192*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 - 264*(c + d*x)*Cos[2*(c + d*
x)] - 16*Sin[c + d*x] + 225*Sin[2*(c + d*x)] - 72*Sin[3*(c + d*x)] + 18*Sin[4*(c + d*x)] + 8*Sin[5*(c + d*x)]
+ Sin[6*(c + d*x)]))/(128*d)

________________________________________________________________________________________

fricas [A]  time = 0.74, size = 152, normalized size = 1.10 \[ -\frac {33 \, a^{3} d x \cos \left (d x + c\right )^{2} - 6 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{3} \cos \left (d x + c\right )^{4} + 7 \, a^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} \cos \left (d x + c\right )^{2} + 24 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(33*a^3*d*x*cos(d*x + c)^2 - 6*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) + 6*a^3*cos(d*x + c)^2*log(-sin(d
*x + c) + 1) - (2*a^3*cos(d*x + c)^5 + 8*a^3*cos(d*x + c)^4 + 7*a^3*cos(d*x + c)^3 - 24*a^3*cos(d*x + c)^2 + 2
4*a^3*cos(d*x + c) + 4*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

giac [A]  time = 0.40, size = 180, normalized size = 1.30 \[ -\frac {33 \, {\left (d x + c\right )} a^{3} - 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {8 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {2 \, {\left (25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 81 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 79 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(33*(d*x + c)*a^3 - 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1))
+ 8*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(25*a^3*ta
n(1/2*d*x + 1/2*c)^7 + 81*a^3*tan(1/2*d*x + 1/2*c)^5 + 79*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*a^3*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.76, size = 159, normalized size = 1.15 \[ \frac {11 a^{3} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}+\frac {33 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {33 a^{3} x}{8}-\frac {33 a^{3} c}{8 d}-\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{3} \sin \left (d x +c \right )}{2 d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*sin(d*x+c)^4,x)

[Out]

11/4*a^3*cos(d*x+c)*sin(d*x+c)^3/d+33/8*a^3*cos(d*x+c)*sin(d*x+c)/d-33/8*a^3*x-33/8/d*a^3*c-1/2*a^3*sin(d*x+c)
^3/d-3/2*a^3*sin(d*x+c)/d+3/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^3*sin(d*x+c)^5/cos(d*x+c)+1/2/d*a^3*sin(d*
x+c)^5/cos(d*x+c)^2

________________________________________________________________________________________

maxima [A]  time = 0.78, size = 182, normalized size = 1.32 \[ -\frac {16 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{3} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{3} + 8 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(16*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*a^3 - (12*d*
x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a^3 + 48*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) -
2*tan(d*x + c))*a^3 + 8*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c
) - 1) - 4*sin(d*x + c)))/d

________________________________________________________________________________________

mupad [B]  time = 1.97, size = 204, normalized size = 1.48 \[ \frac {3\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {33\,a^3\,x}{8}+\frac {-\frac {45\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{4}-\frac {83\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {25\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+\frac {79\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+\frac {27\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {21\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (33*a^3*x)/8 + ((27*a^3*tan(c/2 + (d*x)/2)^3)/4 + (79*a^3*tan(c/2 + (d*x
)/2)^5)/2 + (25*a^3*tan(c/2 + (d*x)/2)^7)/2 - (83*a^3*tan(c/2 + (d*x)/2)^9)/4 - (45*a^3*tan(c/2 + (d*x)/2)^11)
/4 + (21*a^3*tan(c/2 + (d*x)/2))/4)/(d*(2*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^6
- tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________