\(\int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx\) [194]

Optimal result

Integrand size = 21, antiderivative size = 205 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}+\frac {5 b^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {a^3 \cot (c+d x)}{d}-\frac {6 a b^2 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {a b^2 \cot ^3(c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {5 b^3 \csc (c+d x)}{2 d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {5 b^3 \csc ^3(c+d x)}{6 d}+\frac {b^3 \csc ^3(c+d x) \sec ^2(c+d x)}{2 d}+\frac {3 a b^2 \tan (c+d x)}{d} \]

[Out]

3*a^2*b*arctanh(sin(d*x+c))/d+5/2*b^3*arctanh(sin(d*x+c))/d-a^3*cot(d*x+c)/d-6*a*b^2*cot(d*x+c)/d-1/3*a^3*cot(
d*x+c)^3/d-a*b^2*cot(d*x+c)^3/d-3*a^2*b*csc(d*x+c)/d-5/2*b^3*csc(d*x+c)/d-a^2*b*csc(d*x+c)^3/d-5/6*b^3*csc(d*x
+c)^3/d+1/2*b^3*csc(d*x+c)^3*sec(d*x+c)^2/d+3*a*b^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3957, 2991, 3852, 2701, 308, 213, 2700, 276, 294} \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 b \csc ^3(c+d x)}{d}-\frac {3 a^2 b \csc (c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d}-\frac {a b^2 \cot ^3(c+d x)}{d}-\frac {6 a b^2 \cot (c+d x)}{d}+\frac {5 b^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 b^3 \csc ^3(c+d x)}{6 d}-\frac {5 b^3 \csc (c+d x)}{2 d}+\frac {b^3 \csc ^3(c+d x) \sec ^2(c+d x)}{2 d} \]

[In]

Int[Csc[c + d*x]^4*(a + b*Sec[c + d*x])^3,x]

[Out]

(3*a^2*b*ArcTanh[Sin[c + d*x]])/d + (5*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cot[c + d*x])/d - (6*a*b^2*Cot[
c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) - (a*b^2*Cot[c + d*x]^3)/d - (3*a^2*b*Csc[c + d*x])/d - (5*b^3*Csc[c
+ d*x])/(2*d) - (a^2*b*Csc[c + d*x]^3)/d - (5*b^3*Csc[c + d*x]^3)/(6*d) + (b^3*Csc[c + d*x]^3*Sec[c + d*x]^2)/
(2*d) + (3*a*b^2*Tan[c + d*x])/d

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 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 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 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]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(610\) vs. \(2(205)=410\).

Time = 1.44 (sec) , antiderivative size = 610, normalized size of antiderivative = 2.98 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {\csc ^7\left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (84 a^2 b+22 b^3+32 a \left (a^2+3 b^2\right ) \cos (c+d x)+8 \left (6 a^2 b+5 b^3\right ) \cos (2 (c+d x))+4 a^3 \cos (3 (c+d x))+48 a b^2 \cos (3 (c+d x))-36 a^2 b \cos (4 (c+d x))-30 b^3 \cos (4 (c+d x))-4 a^3 \cos (5 (c+d x))-48 a b^2 \cos (5 (c+d x))+36 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+30 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-36 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-30 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+18 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+15 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-18 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-15 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-18 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))-15 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+18 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+15 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{768 d \left (-1+\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )^2} \]

[In]

Integrate[Csc[c + d*x]^4*(a + b*Sec[c + d*x])^3,x]

[Out]

-1/768*(Csc[(c + d*x)/2]^7*Sec[(c + d*x)/2]^3*(84*a^2*b + 22*b^3 + 32*a*(a^2 + 3*b^2)*Cos[c + d*x] + 8*(6*a^2*
b + 5*b^3)*Cos[2*(c + d*x)] + 4*a^3*Cos[3*(c + d*x)] + 48*a*b^2*Cos[3*(c + d*x)] - 36*a^2*b*Cos[4*(c + d*x)] -
30*b^3*Cos[4*(c + d*x)] - 4*a^3*Cos[5*(c + d*x)] - 48*a*b^2*Cos[5*(c + d*x)] + 36*a^2*b*Log[Cos[(c + d*x)/2]
- Sin[(c + d*x)/2]]*Sin[c + d*x] + 30*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 36*a^2*b*Log
[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 30*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d
*x] + 18*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 15*b^3*Log[Cos[(c + d*x)/2] - Sin[(
c + d*x)/2]]*Sin[3*(c + d*x)] - 18*a^2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 15*b^3*Lo
g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 18*a^2*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Si
n[5*(c + d*x)] - 15*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 18*a^2*b*Log[Cos[(c + d*x)
/2] + Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 15*b^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/
(d*(-1 + Cot[(c + d*x)/2]^2)^2)

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+3 a^{2} b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+b^{3} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {5}{6 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {5}{2 \sin \left (d x +c \right )}+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(187\)
default	\(\frac {a^{3} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+3 a^{2} b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )+b^{3} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {5}{6 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {5}{2 \sin \left (d x +c \right )}+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(187\)
parallelrisch	\(\frac {-576 \left (1+\cos \left (2 d x +2 c \right )\right ) b \left (a^{2}+\frac {5 b^{2}}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+576 \left (1+\cos \left (2 d x +2 c \right )\right ) b \left (a^{2}+\frac {5 b^{2}}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (\left (-12 a^{2} b -10 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-a^{3}-12 a \,b^{2}\right ) \cos \left (3 d x +3 c \right )+\left (9 a^{2} b +\frac {15}{2} b^{3}\right ) \cos \left (4 d x +4 c \right )+\left (a^{3}+12 a \,b^{2}\right ) \cos \left (5 d x +5 c \right )+\left (-8 a^{3}-24 a \,b^{2}\right ) \cos \left (d x +c \right )-21 a^{2} b -\frac {11 b^{3}}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{192 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(231\)
norman	\(\frac {-\frac {a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}{24 d}+\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{24 d}+\frac {\left (7 a^{3}-39 a^{2} b +57 a \,b^{2}-25 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{24 d}-\frac {\left (7 a^{3}+39 a^{2} b +57 a \,b^{2}+25 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{24 d}-\frac {\left (13 a^{3}-21 a^{2} b +165 a \,b^{2}-25 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{12 d}+\frac {\left (13 a^{3}+21 a^{2} b +165 a \,b^{2}+25 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{12 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {b \left (6 a^{2}+5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (6 a^{2}+5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(305\)
risch	\(-\frac {i \left (18 a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-20 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-12 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-84 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-22 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-20 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-96 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-20 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-4 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-48 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+18 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+15 b^{3} {\mathrm e}^{i \left (d x +c \right )}+4 a^{3}+48 a \,b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}-\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}+\frac {5 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}\)	\(349\)

[In]

int(csc(d*x+c)^4*(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(-2/3-1/3*csc(d*x+c)^2)*cot(d*x+c)+3*a^2*b*(-1/3/sin(d*x+c)^3-1/sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))
+3*a*b^2*(-1/3/sin(d*x+c)^3/cos(d*x+c)+4/3/sin(d*x+c)/cos(d*x+c)-8/3*cot(d*x+c))+b^3*(-1/3/sin(d*x+c)^3/cos(d*
x+c)^2+5/6/sin(d*x+c)/cos(d*x+c)^2-5/2/sin(d*x+c)+5/2*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.27 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {8 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 36 \, a b^{2} \cos \left (d x + c\right ) - 12 \, {\left (a^{3} + 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, b^{3} - 8 \, {\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left ({\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - {\left (6 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(8*(a^3 + 12*a*b^2)*cos(d*x + c)^5 + 6*(6*a^2*b + 5*b^3)*cos(d*x + c)^4 + 36*a*b^2*cos(d*x + c) - 12*(a^
3 + 12*a*b^2)*cos(d*x + c)^3 + 6*b^3 - 8*(6*a^2*b + 5*b^3)*cos(d*x + c)^2 - 3*((6*a^2*b + 5*b^3)*cos(d*x + c)^
4 - (6*a^2*b + 5*b^3)*cos(d*x + c)^2)*log(sin(d*x + c) + 1)*sin(d*x + c) + 3*((6*a^2*b + 5*b^3)*cos(d*x + c)^4
- (6*a^2*b + 5*b^3)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)*sin(d*x + c))/((d*cos(d*x + c)^4 - d*cos(d*x + c)^
2)*sin(d*x + c))

Sympy [F]

\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \csc ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**4*(a+b*sec(d*x+c))**3,x)

[Out]

Integral((a + b*sec(c + d*x))**3*csc(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.93 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=-\frac {b^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{3}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a b^{2} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{3}}{\tan \left (d x + c\right )^{3}}}{12 \, d} \]

[In]

integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/12*(b^3*(2*(15*sin(d*x + c)^4 - 10*sin(d*x + c)^2 - 2)/(sin(d*x + c)^5 - sin(d*x + c)^3) - 15*log(sin(d*x +
c) + 1) + 15*log(sin(d*x + c) - 1)) + 6*a^2*b*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) +
1) + 3*log(sin(d*x + c) - 1)) + 12*a*b^2*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)) + 4*(3*tan(
d*x + c)^2 + 1)*a^3/tan(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.76 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 63 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, {\left (6 \, a^{2} b + 5 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (6 \, a^{2} b + 5 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {24 \, {\left (6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 63 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 27 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1
/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c) - 45*a^2*b*tan(1/2*d*x + 1/2*c) + 63*a*b^2*tan(1/2*d*x + 1/2*c)
- 27*b^3*tan(1/2*d*x + 1/2*c) + 12*(6*a^2*b + 5*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*(6*a^2*b + 5*b^3
)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 24*(6*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b
^2*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - (9*a^3*tan(1/2*d*x + 1/2*
c)^2 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 63*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 27*b^3*tan(1/2*d*x + 1/2*c)^2 + a^3
+ 3*a^2*b + 3*a*b^2 + b^3)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 13.85 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.27 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\left (a-b\right )}^3}{24\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,b\,{\left (a-b\right )}^2}{4}-\frac {3\,{\left (a-b\right )}^3}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {7\,a^3}{3}+13\,a^2\,b+19\,a\,b^2+\frac {25\,b^3}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {17\,a^3}{3}+29\,a^2\,b+89\,a\,b^2+\frac {77\,b^3}{3}\right )+a\,b^2+a^2\,b+\frac {a^3}{3}+\frac {b^3}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a^3+15\,a^2\,b+69\,a\,b^2+b^3\right )}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2\,b\,6{}\mathrm {i}+b^3\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{d} \]

[In]

int((a + b/cos(c + d*x))^3/sin(c + d*x)^4,x)

[Out]

(tan(c/2 + (d*x)/2)^3*(a - b)^3)/(24*d) - (tan(c/2 + (d*x)/2)*((3*b*(a - b)^2)/4 - (3*(a - b)^3)/8))/d - (atan
h(tan(c/2 + (d*x)/2))*(a^2*b*6i + b^3*5i)*1i)/d - (tan(c/2 + (d*x)/2)^2*(19*a*b^2 + 13*a^2*b + (7*a^3)/3 + (25
*b^3)/3) - tan(c/2 + (d*x)/2)^4*(89*a*b^2 + 29*a^2*b + (17*a^3)/3 + (77*b^3)/3) + a*b^2 + a^2*b + a^3/3 + b^3/
3 + tan(c/2 + (d*x)/2)^6*(69*a*b^2 + 15*a^2*b + 3*a^3 + b^3))/(d*(8*tan(c/2 + (d*x)/2)^3 - 16*tan(c/2 + (d*x)/
2)^5 + 8*tan(c/2 + (d*x)/2)^7))