\(\int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\) [815]

Optimal result

Integrand size = 27, antiderivative size = 72 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {5 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2951, 3855, 2729, 2727} \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {5 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2} \]

[In]

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

[Out]

-((a^3*ArcTanh[Cos[c + d*x]])/d) + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])^2) + (5*a^3*Cos[c + d*x])/(3*d
*(1 - Sin[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2951

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 3855

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

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

Mathematica [A] (verified)

Time = 6.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.00 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 (1+\sin (c+d x))^3 \left (-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (-7+5 \sin (c+d x))}{\left (-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

[In]

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

[Out]

(a^3*(1 + Sin[c + d*x])^3*(-3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2]] + 2/(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^2 + (2*Sin[(c + d*x)/2]*(-7 + 5*Sin[c + d*x]))/(-Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3))/(3*d*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2])^6)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99

method	result	size
parallelrisch	\(\frac {\left (\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {14}{3}\right ) a^{3}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(71\)
risch	\(\frac {2 a^{3} \left (-12 i {\mathrm e}^{i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}-5\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(88\)
derivativedivides	\(\frac {\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a^{3}}{\cos \left (d x +c \right )^{3}}-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(101\)
default	\(\frac {\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a^{3}}{\cos \left (d x +c \right )^{3}}-3 a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{3} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(101\)
norman	\(\frac {-\frac {14 a^{3}}{3 d}-\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {10 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {50 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {20 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {68 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {20 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {50 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {10 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {26 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(266\)

[In]

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

[Out]

((tan(1/2*d*x+1/2*c)-1)^3*ln(tan(1/2*d*x+1/2*c))-6*tan(1/2*d*x+1/2*c)^2+8*tan(1/2*d*x+1/2*c)-14/3)*a^3/d/(tan(
1/2*d*x+1/2*c)-1)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (64) = 128\).

Time = 0.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.21 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {10 \, a^{3} \cos \left (d x + c\right )^{2} + 14 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3} + 3 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (5 \, a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

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

[Out]

-1/6*(10*a^3*cos(d*x + c)^2 + 14*a^3*cos(d*x + c) + 4*a^3 + 3*(a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3 +
(a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a^3*cos(d*x + c)^2 - a^3*cos(d*x +
c) - 2*a^3 + (a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(5*a^3*cos(d*x + c) - 2
*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.43 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 \, a^{3} \tan \left (d x + c\right )^{3} + 6 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} + a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {6 \, a^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

1/3*(3*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(9*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x + 1/2*c) + 7*
a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 10.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-8\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {14\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

[In]

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

[Out]

(a^3*log(tan(c/2 + (d*x)/2)))/d - (6*a^3*tan(c/2 + (d*x)/2)^2 + (14*a^3)/3 - 8*a^3*tan(c/2 + (d*x)/2))/(d*(3*t
an(c/2 + (d*x)/2) - 3*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^3 - 1))