\(\int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx\) [52]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {9 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

9/2*a^3*arctanh(sin(d*x+c))/d-4*a^3*sin(d*x+c)/d/(1-cos(d*x+c))+3*a^3*tan(d*x+c)/d+1/2*a^3*sec(d*x+c)*tan(d*x+
c)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3957, 2951, 2727, 3855, 3852, 8, 3853} \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {9 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}-\frac {4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac {a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 3852

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 3853

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 3855

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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*} \text {integral}& = -\int (-a-a \cos (c+d x))^3 \csc ^2(c+d x) \sec ^3(c+d x) \, dx \\ & = a^2 \int \left (\frac {4 a}{1-\cos (c+d x)}+4 a \sec (c+d x)+3 a \sec ^2(c+d x)+a \sec ^3(c+d x)\right ) \, dx \\ & = a^3 \int \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx+\left (4 a^3\right ) \int \frac {1}{1-\cos (c+d x)} \, dx+\left (4 a^3\right ) \int \sec (c+d x) \, dx \\ & = \frac {4 a^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^3 \int \sec (c+d x) \, dx-\frac {\left (3 a^3\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = \frac {9 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\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 [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(244\) vs. \(2(80)=160\).

Time = 1.52 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.05 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (-18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+16 \csc \left (\frac {c}{2}\right ) \csc \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )+\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{32 d} \]

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(-18*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 18*Log[Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]] + 16*Csc[c/2]*Csc[(c + d*x)/2]*Sin[(d*x)/2] + (Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
^(-2) - (Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(-2) + (12*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]
)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(32*d)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(\frac {9 a^{3} \left (\left (-\cos \left (2 d x +2 c \right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {10 \left (\cos \left (d x +c \right )-\frac {7 \cos \left (2 d x +2 c \right )}{5}-\frac {6}{5}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{9}\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(103\)
norman	\(\frac {-\frac {4 a^{3}}{d}+\frac {15 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {9 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {9 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {9 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(116\)
risch	\(-\frac {i a^{3} \left (9 \,{\mathrm e}^{4 i \left (d x +c \right )}-7 \,{\mathrm e}^{3 i \left (d x +c \right )}+21 \,{\mathrm e}^{2 i \left (d x +c \right )}-5 \,{\mathrm e}^{i \left (d x +c \right )}+14\right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {9 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {9 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(125\)
derivativedivides	\(\frac {a^{3} \left (\frac {1}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 a^{3} \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{3} \cot \left (d x +c \right )}{d}\)	\(127\)
default	\(\frac {a^{3} \left (\frac {1}{2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 a^{3} \left (-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{3} \cot \left (d x +c \right )}{d}\)	\(127\)

[In]

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

[Out]

9/2*a^3*((-cos(2*d*x+2*c)-1)*ln(tan(1/2*d*x+1/2*c)-1)+(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)+10/9*(cos(d*
x+c)-7/5*cos(2*d*x+2*c)-6/5)*cot(1/2*d*x+1/2*c))/d/(1+cos(2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.52 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {9 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 9 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 28 \, a^{3} \cos \left (d x + c\right )^{3} - 18 \, a^{3} \cos \left (d x + c\right )^{2} + 12 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}}{4 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \]

[In]

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

[Out]

1/4*(9*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1)*sin(d*x + c) - 9*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1)*sin
(d*x + c) - 28*a^3*cos(d*x + c)^3 - 18*a^3*cos(d*x + c)^2 + 12*a^3*cos(d*x + c) + 2*a^3)/(d*cos(d*x + c)^2*sin
(d*x + c))

Sympy [F]

\[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 3 \csc ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \csc ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**3*(Integral(3*csc(c + d*x)**2*sec(c + d*x), x) + Integral(3*csc(c + d*x)**2*sec(c + d*x)**2, x) + Integral(
csc(c + d*x)**2*sec(c + d*x)**3, x) + Integral(csc(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.71 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\frac {2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac {4 \, a^{3}}{\tan \left (d x + c\right )}}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.32 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {9 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {8 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(9*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 8*a^3/tan(1/2*d*x +
1/2*c) - 2*(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)/d

Mupad [B] (verification not implemented)

Time = 14.62 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.22 \[ \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {9\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {9\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^3}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

[In]

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

[Out]

(9*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (9*a^3*tan(c/2 + (d*x)/2)^4 - 15*a^3*tan(c/2 + (d*x)/2)^2 + 4*a^3)/(d*(t
an(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^5))