\(\int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1460]

Optimal result

Integrand size = 27, antiderivative size = 78 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-b^3 x-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x)}{d}+\frac {b^3 \tan (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2991, 3852, 8, 2702, 327, 213, 2686, 3554} \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a^2 b \tan (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {b^3 \tan (c+d x)}{d}-b^3 x \]

[In]

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

[Out]

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

Rule 8

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 2702

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

Rule 2991

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

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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]

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

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-b^3 c-b^3 d x-a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a \left (a^2+3 b^2\right ) \sec (c+d x)+b \left (3 a^2+b^2\right ) \tan (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{2} b \tan \left (d x +c \right )+\frac {3 a \,b^{2}}{\cos \left (d x +c \right )}+b^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(79\)
default	\(\frac {a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{2} b \tan \left (d x +c \right )+\frac {3 a \,b^{2}}{\cos \left (d x +c \right )}+b^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(79\)
parallelrisch	\(\frac {a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3} d +2 \left (-3 a^{2} b -b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+b^{3} d x -2 a^{3}-6 a \,b^{2}}{d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(106\)
risch	\(-b^{3} x +\frac {2 i \left (-i a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+3 a^{2} b +b^{3}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(107\)
norman	\(\frac {b^{3} x -\frac {2 a^{3}+6 a \,b^{2}}{d}+2 b^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (a^{3}+3 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (3 a^{3}+9 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {6 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(306\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {2 \, b^{3} d x \cos \left (d x + c\right ) + a^{3} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - a^{3} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, a^{3} - 6 \, a b^{2} - 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} b^{3} - a^{3} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{2} b \tan \left (d x + c\right ) - \frac {6 \, a b^{2}}{\cos \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {{\left (d x + c\right )} b^{3} - a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.97 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b+2\,b^3\right )+6\,a\,b^2+2\,a^3}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {2\,b^3\,\mathrm {atan}\left (\frac {4\,b^6}{4\,a^3\,b^3+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}-\frac {4\,a^3\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,b^3+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}\right )}{d} \]

[In]

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

[Out]

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