\(\int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) [1491]

Optimal result

Integrand size = 27, antiderivative size = 155 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {35 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {35 a \csc (c+d x)}{8 d}-\frac {35 a \csc ^3(c+d x)}{24 d}+\frac {3 b \log (\tan (c+d x))}{d}+\frac {7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 b \tan ^2(c+d x)}{2 d}+\frac {b \tan ^4(c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 155, 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 = {2913, 2701, 294, 308, 213, 2700, 272, 45} \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {35 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {35 a \csc ^3(c+d x)}{24 d}-\frac {35 a \csc (c+d x)}{8 d}+\frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac {7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \tan ^4(c+d x)}{4 d}+\frac {3 b \tan ^2(c+d x)}{2 d}-\frac {b \cot ^2(c+d x)}{2 d}+\frac {3 b \log (\tan (c+d x))}{d} \]

[In]

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

[Out]

(35*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Cot[c + d*x]^2)/(2*d) - (35*a*Csc[c + d*x])/(8*d) - (35*a*Csc[c + d*x]
^3)/(24*d) + (3*b*Log[Tan[c + d*x]])/d + (7*a*Csc[c + d*x]^3*Sec[c + d*x]^2)/(8*d) + (a*Csc[c + d*x]^3*Sec[c +
d*x]^4)/(4*d) + (3*b*Tan[c + d*x]^2)/(2*d) + (b*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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 ^4(c+d x) \sec ^5(c+d x) \, dx+b \int \csc ^3(c+d x) \sec ^5(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int \frac {x^8}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d}+\frac {b \text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^3} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}-\frac {(7 a) \text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d}+\frac {b \text {Subst}\left (\int \frac {(1+x)^3}{x^2} \, dx,x,\tan ^2(c+d x)\right )}{2 d} \\ & = \frac {7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}-\frac {(35 a) \text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}+\frac {b \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 {b \cot ^2(c+d x)}{2 d}+\frac {3 b \log (\tan (c+d x))}{d}+\frac {7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 b \tan ^2(c+d x)}{2 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {(35 a) \text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{8 d} \\ & = -\frac {b \cot ^2(c+d x)}{2 d}-\frac {35 a \csc (c+d x)}{8 d}-\frac {35 a \csc ^3(c+d x)}{24 d}+\frac {3 b \log (\tan (c+d x))}{d}+\frac {7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 b \tan ^2(c+d x)}{2 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {(35 a) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d} \\ & = \frac {35 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \cot ^2(c+d x)}{2 d}-\frac {35 a \csc (c+d x)}{8 d}-\frac {35 a \csc ^3(c+d x)}{24 d}+\frac {3 b \log (\tan (c+d x))}{d}+\frac {7 a \csc ^3(c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc ^3(c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 b \tan ^2(c+d x)}{2 d}+\frac {b \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 = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.67 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \csc ^2(c+d x)}{2 d}-\frac {a \csc ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},\sin ^2(c+d x)\right )}{3 d}-\frac {3 b \log (\cos (c+d x))}{d}+\frac {3 b \log (\sin (c+d x))}{d}+\frac {b \sec ^2(c+d x)}{d}+\frac {b \sec ^4(c+d x)}{4 d} \]

[In]

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

[Out]

-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*L
og[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)

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95

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

[In]

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

[Out]

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)

none

Time = 0.32 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.60 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {210 \, a \cos \left (d x + c\right )^{6} - 280 \, a \cos \left (d x + c\right )^{4} + 42 \, a \cos \left (d x + c\right )^{2} - 144 \, {\left (b \cos \left (d x + c\right )^{6} - b \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 3 \, {\left ({\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a - 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a + 24 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 12 \, {\left (6 \, b \cos \left (d x + c\right )^{4} - 3 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) + 12 \, a}{48 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, {\left (35 \, a - 24 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (35 \, a + 24 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (105 \, a \sin \left (d x + c\right )^{6} + 36 \, b \sin \left (d x + c\right )^{5} - 175 \, a \sin \left (d x + c\right )^{4} - 54 \, b \sin \left (d x + c\right )^{3} + 56 \, a \sin \left (d x + c\right )^{2} + 12 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{7} - 2 \, \sin \left (d x + c\right )^{5} + \sin \left (d x + c\right )^{3}}}{48 \, d} \]

[In]

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

[Out]

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)

none

Time = 0.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, {\left (35 \, a - 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (35 \, a + 24 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 144 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {6 \, {\left (18 \, b \sin \left (d x + c\right )^{4} - 11 \, a \sin \left (d x + c\right )^{3} - 44 \, b \sin \left (d x + c\right )^{2} + 13 \, a \sin \left (d x + c\right ) + 28 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}} - \frac {8 \, {\left (33 \, b \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} + 3 \, b \sin \left (d x + c\right ) + 2 \, a\right )}}{\sin \left (d x + c\right )^{3}}}{48 \, d} \]

[In]

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

[Out]

1/48*(3*(35*a - 24*b)*log(abs(sin(d*x + c) + 1)) - 3*(35*a + 24*b)*log(abs(sin(d*x + c) - 1)) + 144*b*log(abs(
sin(d*x + c))) + 6*(18*b*sin(d*x + c)^4 - 11*a*sin(d*x + c)^3 - 44*b*sin(d*x + c)^2 + 13*a*sin(d*x + c) + 28*b
)/(sin(d*x + c)^2 - 1)^2 - 8*(33*b*sin(d*x + c)^3 + 18*a*sin(d*x + c)^2 + 3*b*sin(d*x + c) + 2*a)/sin(d*x + c)
^3)/d

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {35\,a}{16}-\frac {3\,b}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {35\,a}{16}+\frac {3\,b}{2}\right )}{d}+\frac {3\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {35\,a\,{\sin \left (c+d\,x\right )}^6}{8}+\frac {3\,b\,{\sin \left (c+d\,x\right )}^5}{2}-\frac {175\,a\,{\sin \left (c+d\,x\right )}^4}{24}-\frac {9\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {7\,a\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {b\,\sin \left (c+d\,x\right )}{2}+\frac {a}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^7-2\,{\sin \left (c+d\,x\right )}^5+{\sin \left (c+d\,x\right )}^3\right )} \]

[In]

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

[Out]

(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(s
in(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))