\(\int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx\) [35]

Optimal result

Integrand size = 19, antiderivative size = 86 \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}-\frac {b^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3598, 3855, 2686, 8, 2691} \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {b^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[In]

Int[Csc[c + d*x]*(a + b*Tan[c + d*x])^3,x]

[Out]

-((a^3*ArcTanh[Cos[c + d*x]])/d) + (3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (b^3*ArcTanh[Sin[c + d*x]])/(2*d) + (3*
a*b^2*Sec[c + d*x])/d + (b^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 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3598

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 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}& = \int \left (a^3 \csc (c+d x)+3 a^2 b \sec (c+d x)+3 a b^2 \sec (c+d x) \tan (c+d x)+b^3 \sec (c+d x) \tan ^2(c+d x)\right ) \, dx \\ & = a^3 \int \csc (c+d x) \, dx+\left (3 a^2 b\right ) \int \sec (c+d x) \, dx+\left (3 a b^2\right ) \int \sec (c+d x) \tan (c+d x) \, dx+b^3 \int \sec (c+d x) \tan ^2(c+d x) \, dx \\ & = -\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}+\frac {b^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} b^3 \int \sec (c+d x) \, dx+\frac {\left (3 a b^2\right ) \text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d} \\ & = -\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}-\frac {b^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^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. \(241\) vs. \(2(86)=172\).

Time = 3.87 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.80 \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {12 a b^2-4 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-12 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+24 a b^2 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-\frac {b^3}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}}{4 d} \]

[In]

Integrate[Csc[c + d*x]*(a + b*Tan[c + d*x])^3,x]

[Out]

(12*a*b^2 - 4*a^3*Log[Cos[(c + d*x)/2]] - 12*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*b^3*Log[Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]] + 4*a^3*Log[Sin[(c + d*x)/2]] + 12*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]] - 2*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^3/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 24*a*b^2*S
ec[c + d*x]*Sin[(c + d*x)/2]^2 - b^3/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(4*d)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 a \,b^{2}}{\cos \left (d x +c \right )}+3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(107\)
default	\(\frac {b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 a \,b^{2}}{\cos \left (d x +c \right )}+3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(107\)
risch	\(-\frac {i b^{2} {\mathrm e}^{i \left (d x +c \right )} \left (6 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 i a -b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(196\)

[In]

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

[Out]

1/d*(b^3*(1/2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+tan(d*x+c)))+3*a*b^2/cos(d*x+c)+3*a^2
*b*ln(sec(d*x+c)+tan(d*x+c))+a^3*ln(csc(d*x+c)-cot(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.72 \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, a b^{2} \cos \left (d x + c\right ) - {\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, b^{3} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-1/4*(2*a^3*cos(d*x + c)^2*log(1/2*cos(d*x + c) + 1/2) - 2*a^3*cos(d*x + c)^2*log(-1/2*cos(d*x + c) + 1/2) - 1
2*a*b^2*cos(d*x + c) - (6*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (6*a^2*b - b^3)*cos(d*x + c)^2*l
og(-sin(d*x + c) + 1) - 2*b^3*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F]

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

[In]

integrate(csc(d*x+c)*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*csc(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29 \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, a^{3} \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right ) - \frac {12 \, a b^{2}}{\cos \left (d x + c\right )}}{4 \, d} \]

[In]

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

[Out]

-1/4*(b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*a^2*b*(log
(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*a^3*log(cot(d*x + c) + csc(d*x + c)) - 12*a*b^2/cos(d*x + c))/
d

Giac [A] (verification not implemented)

none

Time = 0.68 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.67 \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + {\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.23 \[ \int \csc (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2\,\left (\frac {a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-\frac {b^3\,\mathrm {atan}\left (\frac {2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3-6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{2{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3-6{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b+1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,1{}\mathrm {i}}{2}+a^2\,b\,\mathrm {atan}\left (\frac {2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3-6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}{2{}\mathrm {i}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3-6{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b+1{}\mathrm {i}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^3}\right )\,3{}\mathrm {i}\right )}{d}+\frac {\frac {\sin \left (c+d\,x\right )\,b^3}{2}+3\,a\,\cos \left (c+d\,x\right )\,b^2}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

[In]

int((a + b*tan(c + d*x))^3/sin(c + d*x),x)

[Out]

(2*((a^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (b^3*atan((b^3*cos(c/2 + (d*x)/2) + 2*a^3*sin(c/2 + (
d*x)/2) - 6*a^2*b*cos(c/2 + (d*x)/2))/(a^3*cos(c/2 + (d*x)/2)*2i + b^3*sin(c/2 + (d*x)/2)*1i - a^2*b*sin(c/2 +
(d*x)/2)*6i))*1i)/2 + a^2*b*atan((b^3*cos(c/2 + (d*x)/2) + 2*a^3*sin(c/2 + (d*x)/2) - 6*a^2*b*cos(c/2 + (d*x)
/2))/(a^3*cos(c/2 + (d*x)/2)*2i + b^3*sin(c/2 + (d*x)/2)*1i - a^2*b*sin(c/2 + (d*x)/2)*6i))*3i))/d + ((b^3*sin
(c + d*x))/2 + 3*a*b^2*cos(c + d*x))/(d*(cos(2*c + 2*d*x)/2 + 1/2))