\(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [1490]

Optimal result

Integrand size = 27, antiderivative size = 135 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \cot ^2(c+d x)}{2 d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {3 a \log (\tan (c+d x))}{d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}+\frac {a \tan ^4(c+d x)}{4 d} \]

[Out]

15/8*b*arctanh(sin(d*x+c))/d-1/2*a*cot(d*x+c)^2/d-15/8*b*csc(d*x+c)/d+3*a*ln(tan(d*x+c))/d+5/8*b*csc(d*x+c)*se
c(d*x+c)^2/d+1/4*b*csc(d*x+c)*sec(d*x+c)^4/d+3/2*a*tan(d*x+c)^2/d+1/4*a*tan(d*x+c)^4/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2913, 2700, 272, 45, 2701, 294, 327, 213} \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \tan ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}-\frac {a \cot ^2(c+d x)}{2 d}+\frac {3 a \log (\tan (c+d x))}{d}+\frac {15 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d} \]

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(15*b*ArcTanh[Sin[c + d*x]])/(8*d) - (a*Cot[c + d*x]^2)/(2*d) - (15*b*Csc[c + d*x])/(8*d) + (3*a*Log[Tan[c + d
*x]])/d + (5*b*Csc[c + d*x]*Sec[c + d*x]^2)/(8*d) + (b*Csc[c + d*x]*Sec[c + d*x]^4)/(4*d) + (3*a*Tan[c + d*x]^
2)/(2*d) + (a*Tan[c + d*x]^4)/(4*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 2700

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

Rule 2701

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

Rule 2913

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

Rubi steps \begin{align*} \text {integral}& = a \int \csc ^3(c+d x) \sec ^5(c+d x) \, dx+b \int \csc ^2(c+d x) \sec ^5(c+d x) \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^3} \, dx,x,\tan (c+d x)\right )}{d}-\frac {b \text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d} \\ & = \frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {a \text {Subst}\left (\int \frac {(1+x)^3}{x^2} \, dx,x,\tan ^2(c+d x)\right )}{2 d}-\frac {(5 b) \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d} \\ & = \frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {a \text {Subst}\left (\int \left (3+\frac {1}{x^2}+\frac {3}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d}-\frac {(15 b) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d} \\ & = -\frac {a \cot ^2(c+d x)}{2 d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {3 a \log (\tan (c+d x))}{d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}+\frac {a \tan ^4(c+d x)}{4 d}-\frac {(15 b) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d} \\ & = \frac {15 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \cot ^2(c+d x)}{2 d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {3 a \log (\tan (c+d x))}{d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}+\frac {a \tan ^4(c+d x)}{4 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.74 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc ^2(c+d x)}{2 d}-\frac {b \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(c+d x)\right )}{d}-\frac {3 a \log (\cos (c+d x))}{d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec ^4(c+d x)}{4 d} \]

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

-1/2*(a*Csc[c + d*x]^2)/d - (b*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[c + d*x]^2])/d - (3*a*Log[Cos[
c + d*x]])/d + (3*a*Log[Sin[c + d*x]])/d + (a*Sec[c + d*x]^2)/d + (a*Sec[c + d*x]^4)/(4*d)

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(129\)
default	\(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(129\)
parallelrisch	\(\frac {-12 \left (a +\frac {5 b}{8}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {5 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {27 \left (a \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (2 d x +2 c \right )+\frac {2 \cos \left (4 d x +4 c \right )}{9}-\frac {\cos \left (6 d x +6 c \right )}{9}+\frac {2}{27}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {80 b \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right )}{27}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(247\)
risch	\(-\frac {i \left (24 i a \,{\mathrm e}^{10 i \left (d x +c \right )}+15 b \,{\mathrm e}^{11 i \left (d x +c \right )}+48 i a \,{\mathrm e}^{8 i \left (d x +c \right )}+25 b \,{\mathrm e}^{9 i \left (d x +c \right )}-16 i a \,{\mathrm e}^{6 i \left (d x +c \right )}-22 b \,{\mathrm e}^{7 i \left (d x +c \right )}+48 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+22 b \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-25 b \,{\mathrm e}^{3 i \left (d x +c \right )}-15 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(266\)
norman	\(\frac {-\frac {a}{8 d}-\frac {a \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {13 b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {b \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {21 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {21 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {27 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {27 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 \left (8 a -5 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {3 \left (8 a +5 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\)	\(317\)

[In]

int(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(1/4/sin(d*x+c)^2/cos(d*x+c)^4+3/4/sin(d*x+c)^2/cos(d*x+c)^2-3/2/sin(d*x+c)^2+3*ln(tan(d*x+c)))+b*(1/4/
sin(d*x+c)/cos(d*x+c)^4+5/8/sin(d*x+c)/cos(d*x+c)^2-15/8/sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.56 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {24 \, a \cos \left (d x + c\right )^{4} - 12 \, a \cos \left (d x + c\right )^{2} + 48 \, {\left (a \cos \left (d x + c\right )^{6} - a \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 3 \, {\left ({\left (8 \, a - 5 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (8 \, a - 5 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (8 \, a + 5 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (8 \, a + 5 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, b \cos \left (d x + c\right )^{4} - 5 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(24*a*cos(d*x + c)^4 - 12*a*cos(d*x + c)^2 + 48*(a*cos(d*x + c)^6 - a*cos(d*x + c)^4)*log(1/2*sin(d*x + c
)) - 3*((8*a - 5*b)*cos(d*x + c)^6 - (8*a - 5*b)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 3*((8*a + 5*b)*cos(d*
x + c)^6 - (8*a + 5*b)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(15*b*cos(d*x + c)^4 - 5*b*cos(d*x + c)^2 -
2*b)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^6 - d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**5*(a+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {3 \, {\left (8 \, a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, a + 5 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, b \sin \left (d x + c\right )^{5} + 12 \, a \sin \left (d x + c\right )^{4} - 25 \, b \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 8 \, b \sin \left (d x + c\right ) + 4 \, a\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{16 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(3*(8*a - 5*b)*log(sin(d*x + c) + 1) + 3*(8*a + 5*b)*log(sin(d*x + c) - 1) - 48*a*log(sin(d*x + c)) + 2*
(15*b*sin(d*x + c)^5 + 12*a*sin(d*x + c)^4 - 25*b*sin(d*x + c)^3 - 18*a*sin(d*x + c)^2 + 8*b*sin(d*x + c) + 4*
a)/(sin(d*x + c)^6 - 2*sin(d*x + c)^4 + sin(d*x + c)^2))/d

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {3 \, {\left (8 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (8 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 48 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (15 \, b \sin \left (d x + c\right )^{5} + 12 \, a \sin \left (d x + c\right )^{4} - 25 \, b \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 8 \, b \sin \left (d x + c\right ) + 4 \, a\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{16 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(3*(8*a - 5*b)*log(abs(sin(d*x + c) + 1)) + 3*(8*a + 5*b)*log(abs(sin(d*x + c) - 1)) - 48*a*log(abs(sin(
d*x + c))) + 2*(15*b*sin(d*x + c)^5 + 12*a*sin(d*x + c)^4 - 25*b*sin(d*x + c)^3 - 18*a*sin(d*x + c)^2 + 8*b*si
n(d*x + c) + 4*a)/(sin(d*x + c)^3 - sin(d*x + c))^2)/d

Mupad [B] (verification not implemented)

Time = 11.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {3\,a}{2}-\frac {15\,b}{16}\right )}{d}-\frac {\frac {15\,b\,{\sin \left (c+d\,x\right )}^5}{8}+\frac {3\,a\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {25\,b\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {9\,a\,{\sin \left (c+d\,x\right )}^2}{4}+b\,\sin \left (c+d\,x\right )+\frac {a}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^6-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,a}{2}+\frac {15\,b}{16}\right )}{d} \]

[In]

int((a + b*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)^3),x)

[Out]

(3*a*log(sin(c + d*x)))/d - (log(sin(c + d*x) + 1)*((3*a)/2 - (15*b)/16))/d - (a/2 + b*sin(c + d*x) - (9*a*sin
(c + d*x)^2)/4 + (3*a*sin(c + d*x)^4)/2 - (25*b*sin(c + d*x)^3)/8 + (15*b*sin(c + d*x)^5)/8)/(d*(sin(c + d*x)^
2 - 2*sin(c + d*x)^4 + sin(c + d*x)^6)) - (log(sin(c + d*x) - 1)*((3*a)/2 + (15*b)/16))/d