\(\int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [1487]

Optimal result

Integrand size = 27, antiderivative size = 147 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {35 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {3 a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}+\frac {3 b \log (\tan (c+d x))}{d}+\frac {13 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d} \] Output:

35/8*a*arctanh(sin(d*x+c))/d-1/2*b*cot(d*x+c)^2/d-3*a*csc(d*x+c)/d-1/3*a*c 
sc(d*x+c)^3/d+3*b*ln(tan(d*x+c))/d+13/8*a*sec(d*x+c)*tan(d*x+c)/d+3/2*b*ta 
n(d*x+c)^2/d+1/4*a*sec(d*x+c)*tan(d*x+c)^3/d+1/4*b*tan(d*x+c)^4/d
 

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.71 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \csc ^2(c+d x)}{2 d}-\frac {a \csc ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},\sin ^2(c+d x)\right )}{3 d}-\frac {3 b \log (\cos (c+d x))}{d}+\frac {3 b \log (\sin (c+d x))}{d}+\frac {b \sec ^2(c+d x)}{d}+\frac {b \sec ^4(c+d x)}{4 d} \] Input:

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3313, 3042, 3100, 243, 49, 2009, 3101, 25, 252, 252, 254, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a+b \sin (c+d x)}{\sin (c+d x)^4 \cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3313

\(\displaystyle a \int \csc ^4(c+d x) \sec ^5(c+d x)dx+b \int \csc ^3(c+d x) \sec ^5(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \csc (c+d x)^4 \sec (c+d x)^5dx+b \int \csc (c+d x)^3 \sec (c+d x)^5dx\)

\(\Big \downarrow \) 3100

\(\displaystyle a \int \csc (c+d x)^4 \sec (c+d x)^5dx+\frac {b \int \cot ^3(c+d x) \left (\tan ^2(c+d x)+1\right )^3d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 243

\(\displaystyle a \int \csc (c+d x)^4 \sec (c+d x)^5dx+\frac {b \int \cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right )^3d\tan ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 49

\(\displaystyle a \int \csc (c+d x)^4 \sec (c+d x)^5dx+\frac {b \int \left (\cot ^2(c+d x)+3 \cot (c+d x)+\tan ^2(c+d x)+3\right )d\tan ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle a \int \csc (c+d x)^4 \sec (c+d x)^5dx+\frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}\)

\(\Big \downarrow \) 3101

\(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \int -\frac {\csc ^8(c+d x)}{\left (1-\csc ^2(c+d x)\right )^3}d\csc (c+d x)}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a \int \frac {\csc ^8(c+d x)}{\left (1-\csc ^2(c+d x)\right )^3}d\csc (c+d x)}{d}+\frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {7}{4} \int \frac {\csc ^6(c+d x)}{\left (1-\csc ^2(c+d x)\right )^2}d\csc (c+d x)-\frac {\csc ^7(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {7}{4} \left (\frac {\csc ^5(c+d x)}{2 \left (1-\csc ^2(c+d x)\right )}-\frac {5}{2} \int \frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)\right )-\frac {\csc ^7(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 254

\(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {7}{4} \left (\frac {\csc ^5(c+d x)}{2 \left (1-\csc ^2(c+d x)\right )}-\frac {5}{2} \int \left (-\csc ^2(c+d x)+\frac {1}{1-\csc ^2(c+d x)}-1\right )d\csc (c+d x)\right )-\frac {\csc ^7(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+3 \tan ^2(c+d x)-\cot (c+d x)+3 \log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {7}{4} \left (\frac {\csc ^5(c+d x)}{2 \left (1-\csc ^2(c+d x)\right )}-\frac {5}{2} \left (\text {arctanh}(\csc (c+d x))-\frac {1}{3} \csc ^3(c+d x)-\csc (c+d x)\right )\right )-\frac {\csc ^7(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\)

Input:

Int[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]
 

Output:

-((a*(-1/4*Csc[c + d*x]^7/(1 - Csc[c + d*x]^2)^2 + (7*(Csc[c + d*x]^5/(2*( 
1 - Csc[c + d*x]^2)) - (5*(ArcTanh[Csc[c + d*x]] - Csc[c + d*x] - Csc[c + 
d*x]^3/3))/2))/4))/d) + (b*(-Cot[c + d*x] + 3*Log[Tan[c + d*x]^2] + 3*Tan[ 
c + d*x]^2 + Tan[c + d*x]^4/2))/(2*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3100

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] 
:> Simp[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 3101

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S 
ymbol] :> Simp[-(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] && IntegerQ[(n 
 + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
 

rule 3313

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[Cos[e + f*x]^ 
p*(d*Sin[e + f*x])^n, x], x] + Simp[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])
 

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7}{12 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {35}{24 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {35}{8 \sin \left (d x +c \right )}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(147\)
default	\(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7}{12 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {35}{24 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {35}{8 \sin \left (d x +c \right )}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(147\)
parallelrisch	\(\frac {-13440 \left (a +\frac {24 b}{35}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+13440 \left (a -\frac {24 b}{35}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+9216 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-329 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (2 d x +2 c \right )-\frac {10 \cos \left (4 d x +4 c \right )}{47}-\frac {15 \cos \left (6 d x +6 c \right )}{47}+\frac {102}{329}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {648 \left (\cos \left (2 d x +2 c \right )+\frac {2 \cos \left (4 d x +4 c \right )}{9}-\frac {\cos \left (6 d x +6 c \right )}{9}+\frac {2}{27}\right ) b}{329}\right )}{768 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(262\)
risch	\(-\frac {i \left (105 a \,{\mathrm e}^{13 i \left (d x +c \right )}+70 a \,{\mathrm e}^{11 i \left (d x +c \right )}+72 i b \,{\mathrm e}^{12 i \left (d x +c \right )}-329 a \,{\mathrm e}^{9 i \left (d x +c \right )}+72 i b \,{\mathrm e}^{10 i \left (d x +c \right )}-204 a \,{\mathrm e}^{7 i \left (d x +c \right )}-192 i b \,{\mathrm e}^{8 i \left (d x +c \right )}-329 a \,{\mathrm e}^{5 i \left (d x +c \right )}+192 i b \,{\mathrm e}^{6 i \left (d x +c \right )}+70 a \,{\mathrm e}^{3 i \left (d x +c \right )}-72 i b \,{\mathrm e}^{4 i \left (d x +c \right )}+105 a \,{\mathrm e}^{i \left (d x +c \right )}-72 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {35 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}-\frac {35 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(291\)

Input:

int(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(a*(1/4/sin(d*x+c)^3/cos(d*x+c)^4-7/12/sin(d*x+c)^3/cos(d*x+c)^2+35/24 
/sin(d*x+c)/cos(d*x+c)^2-35/8/sin(d*x+c)+35/8*ln(sec(d*x+c)+tan(d*x+c)))+b 
*(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))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.69 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {210 \, a \cos \left (d x + c\right )^{6} - 280 \, a \cos \left (d x + c\right )^{4} + 42 \, a \cos \left (d x + c\right )^{2} - 144 \, {\left (b \cos \left (d x + c\right )^{6} - b \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 3 \, {\left ({\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 12 \, {\left (6 \, b \cos \left (d x + c\right )^{4} - 3 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) + 12 \, a}{48 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )} \] Input:

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas" 
)
 

Output:

-1/48*(210*a*cos(d*x + c)^6 - 280*a*cos(d*x + c)^4 + 42*a*cos(d*x + c)^2 - 
 144*(b*cos(d*x + c)^6 - b*cos(d*x + c)^4)*log(1/2*sin(d*x + c))*sin(d*x + 
 c) - 3*((35*a - 24*b)*cos(d*x + c)^6 - (35*a - 24*b)*cos(d*x + c)^4)*log( 
sin(d*x + c) + 1)*sin(d*x + c) + 3*((35*a + 24*b)*cos(d*x + c)^6 - (35*a + 
 24*b)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1)*sin(d*x + c) - 12*(6*b*cos(d 
*x + c)^4 - 3*b*cos(d*x + c)^2 - b)*sin(d*x + c) + 12*a)/((d*cos(d*x + c)^ 
6 - d*cos(d*x + c)^4)*sin(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(csc(d*x+c)**4*sec(d*x+c)**5*(a+b*sin(d*x+c)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, {\left (35 \, a - 24 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, a + 24 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (105 \, a \sin \left (d x + c\right )^{6} + 36 \, b \sin \left (d x + c\right )^{5} - 175 \, a \sin \left (d x + c\right )^{4} - 54 \, b \sin \left (d x + c\right )^{3} + 56 \, a \sin \left (d x + c\right )^{2} + 12 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{7} - 2 \, \sin \left (d x + c\right )^{5} + \sin \left (d x + c\right )^{3}}}{48 \, d} \] Input:

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima" 
)
 

Output:

1/48*(3*(35*a - 24*b)*log(sin(d*x + c) + 1) - 3*(35*a + 24*b)*log(sin(d*x 
+ c) - 1) + 144*b*log(sin(d*x + c)) - 2*(105*a*sin(d*x + c)^6 + 36*b*sin(d 
*x + c)^5 - 175*a*sin(d*x + c)^4 - 54*b*sin(d*x + c)^3 + 56*a*sin(d*x + c) 
^2 + 12*b*sin(d*x + c) + 8*a)/(sin(d*x + c)^7 - 2*sin(d*x + c)^5 + sin(d*x 
 + c)^3))/d
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (35 \, a - 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (35 \, a + 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {3 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{d} - \frac {105 \, a \sin \left (d x + c\right )^{6} + 36 \, b \sin \left (d x + c\right )^{5} - 175 \, a \sin \left (d x + c\right )^{4} - 54 \, b \sin \left (d x + c\right )^{3} + 56 \, a \sin \left (d x + c\right )^{2} + 12 \, b \sin \left (d x + c\right ) + 8 \, a}{24 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2} \sin \left (d x + c\right )^{3}} \] Input:

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/16*(35*a - 24*b)*log(abs(sin(d*x + c) + 1))/d - 1/16*(35*a + 24*b)*log(a 
bs(sin(d*x + c) - 1))/d + 3*b*log(abs(sin(d*x + c)))/d - 1/24*(105*a*sin(d 
*x + c)^6 + 36*b*sin(d*x + c)^5 - 175*a*sin(d*x + c)^4 - 54*b*sin(d*x + c) 
^3 + 56*a*sin(d*x + c)^2 + 12*b*sin(d*x + c) + 8*a)/(d*(sin(d*x + c) + 1)^ 
2*(sin(d*x + c) - 1)^2*sin(d*x + c)^3)
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {35\,a}{16}-\frac {3\,b}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {35\,a}{16}+\frac {3\,b}{2}\right )}{d}+\frac {3\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {35\,a\,{\sin \left (c+d\,x\right )}^6}{8}+\frac {3\,b\,{\sin \left (c+d\,x\right )}^5}{2}-\frac {175\,a\,{\sin \left (c+d\,x\right )}^4}{24}-\frac {9\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {7\,a\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {b\,\sin \left (c+d\,x\right )}{2}+\frac {a}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^7-2\,{\sin \left (c+d\,x\right )}^5+{\sin \left (c+d\,x\right )}^3\right )} \] Input:

int((a + b*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)^4),x)
 

Output:

(log(sin(c + d*x) + 1)*((35*a)/16 - (3*b)/2))/d - (log(sin(c + d*x) - 1)*( 
(35*a)/16 + (3*b)/2))/d + (3*b*log(sin(c + d*x)))/d - (a/3 + (b*sin(c + d* 
x))/2 + (7*a*sin(c + d*x)^2)/3 - (175*a*sin(c + d*x)^4)/24 + (35*a*sin(c + 
 d*x)^6)/8 - (9*b*sin(c + d*x)^3)/4 + (3*b*sin(c + d*x)^5)/2)/(d*(sin(c + 
d*x)^3 - 2*sin(c + d*x)^5 + sin(c + d*x)^7))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 453, normalized size of antiderivative = 3.08 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx =\text {Too large to display} \] Input:

int(csc(d*x+c)^4*sec(d*x+c)^5*(a+b*sin(d*x+c)),x)
 

Output:

( - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7*a - 72*log(tan((c + d*x) 
/2) - 1)*sin(c + d*x)**7*b + 210*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 
*a + 144*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*b - 105*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)**3*a - 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 
3*b + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**7*a - 72*log(tan((c + d* 
x)/2) + 1)*sin(c + d*x)**7*b - 210*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* 
*5*a + 144*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*b + 105*log(tan((c + 
d*x)/2) + 1)*sin(c + d*x)**3*a - 72*log(tan((c + d*x)/2) + 1)*sin(c + d*x) 
**3*b + 72*log(tan((c + d*x)/2))*sin(c + d*x)**7*b - 144*log(tan((c + d*x) 
/2))*sin(c + d*x)**5*b + 72*log(tan((c + d*x)/2))*sin(c + d*x)**3*b - 36*s 
in(c + d*x)**7*b - 105*sin(c + d*x)**6*a + 36*sin(c + d*x)**5*b + 175*sin( 
c + d*x)**4*a + 18*sin(c + d*x)**3*b - 56*sin(c + d*x)**2*a - 12*sin(c + d 
*x)*b - 8*a)/(24*sin(c + d*x)**3*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 
1))