3.88 \(\int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=91 \[ -\frac {2 \cot ^7(c+d x)}{7 a^2 d}-\frac {3 \cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {2 \csc ^7(c+d x)}{7 a^2 d}-\frac {2 \csc ^5(c+d x)}{5 a^2 d} \]

[Out]

-1/3*cot(d*x+c)^3/a^2/d-3/5*cot(d*x+c)^5/a^2/d-2/7*cot(d*x+c)^7/a^2/d-2/5*csc(d*x+c)^5/a^2/d+2/7*csc(d*x+c)^7/
a^2/d

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Rubi [A]  time = 0.34, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3872, 2875, 2873, 2607, 14, 2606, 270} \[ -\frac {2 \cot ^7(c+d x)}{7 a^2 d}-\frac {3 \cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {2 \csc ^7(c+d x)}{7 a^2 d}-\frac {2 \csc ^5(c+d x)}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]

[Out]

-Cot[c + d*x]^3/(3*a^2*d) - (3*Cot[c + d*x]^5)/(5*a^2*d) - (2*Cot[c + d*x]^7)/(7*a^2*d) - (2*Csc[c + d*x]^5)/(
5*a^2*d) + (2*Csc[c + d*x]^7)/(7*a^2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

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 2606

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 2607

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

Rule 2873

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] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + 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] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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*} \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\int (-a+a \cos (c+d x))^2 \cot ^2(c+d x) \csc ^6(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cot ^4(c+d x) \csc ^4(c+d x)-2 a^2 \cot ^3(c+d x) \csc ^5(c+d x)+a^2 \cot ^2(c+d x) \csc ^6(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cot ^4(c+d x) \csc ^4(c+d x) \, dx}{a^2}+\frac {\int \cot ^2(c+d x) \csc ^6(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^3(c+d x) \csc ^5(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int x^2 \left (1+x^2\right )^2 \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {2 \operatorname {Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{a^2 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int \left (x^2+2 x^4+x^6\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}+\frac {2 \operatorname {Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\csc (c+d x)\right )}{a^2 d}\\ &=-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {3 \cot ^5(c+d x)}{5 a^2 d}-\frac {2 \cot ^7(c+d x)}{7 a^2 d}-\frac {2 \csc ^5(c+d x)}{5 a^2 d}+\frac {2 \csc ^7(c+d x)}{7 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.70, size = 149, normalized size = 1.64 \[ -\frac {\csc (c) (-714 \sin (c+d x)-408 \sin (2 (c+d x))+153 \sin (3 (c+d x))+204 \sin (4 (c+d x))+51 \sin (5 (c+d x))+1680 \sin (2 c+d x)+128 \sin (c+2 d x)-48 \sin (2 c+3 d x)-64 \sin (3 c+4 d x)-16 \sin (4 c+5 d x)+1344 \sin (c)-1456 \sin (d x)) \csc ^3(c+d x) \sec ^2(c+d x)}{13440 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]

[Out]

-1/13440*(Csc[c]*Csc[c + d*x]^3*Sec[c + d*x]^2*(1344*Sin[c] - 1456*Sin[d*x] - 714*Sin[c + d*x] - 408*Sin[2*(c
+ d*x)] + 153*Sin[3*(c + d*x)] + 204*Sin[4*(c + d*x)] + 51*Sin[5*(c + d*x)] + 1680*Sin[2*c + d*x] + 128*Sin[c
+ 2*d*x] - 48*Sin[2*c + 3*d*x] - 64*Sin[3*c + 4*d*x] - 16*Sin[4*c + 5*d*x]))/(a^2*d*(1 + Sec[c + d*x])^2)

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fricas [A]  time = 0.79, size = 108, normalized size = 1.19 \[ \frac {2 \, \cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} + 24 \, \cos \left (d x + c\right ) + 12}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/105*(2*cos(d*x + c)^5 + 4*cos(d*x + c)^4 - cos(d*x + c)^3 - 6*cos(d*x + c)^2 + 24*cos(d*x + c) + 12)/((a^2*d
*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))

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giac [A]  time = 0.38, size = 105, normalized size = 1.15 \[ -\frac {\frac {35 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {15 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 70 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 210 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{14}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3360*(35*(3*tan(1/2*d*x + 1/2*c)^2 + 1)/(a^2*tan(1/2*d*x + 1/2*c)^3) - (15*a^12*tan(1/2*d*x + 1/2*c)^7 + 21
*a^12*tan(1/2*d*x + 1/2*c)^5 - 70*a^12*tan(1/2*d*x + 1/2*c)^3 - 210*a^12*tan(1/2*d*x + 1/2*c))/a^14)/d

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maple [A]  time = 0.80, size = 86, normalized size = 0.95 \[ \frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{32 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+a*sec(d*x+c))^2,x)

[Out]

1/32/d/a^2*(1/7*tan(1/2*d*x+1/2*c)^7+1/5*tan(1/2*d*x+1/2*c)^5-2/3*tan(1/2*d*x+1/2*c)^3-2*tan(1/2*d*x+1/2*c)-1/
3/tan(1/2*d*x+1/2*c)^3-1/tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.48, size = 134, normalized size = 1.47 \[ -\frac {\frac {\frac {210 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{2}} + \frac {35 \, {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{2} \sin \left (d x + c\right )^{3}}}{3360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3360*((210*sin(d*x + c)/(cos(d*x + c) + 1) + 70*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^2 + 35*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)*(cos(d*x + c) + 1)^3/(a^2*sin(d*x + c)^3))/d

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mupad [B]  time = 1.07, size = 121, normalized size = 1.33 \[ -\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-96\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+54\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-15}{3360\,a^2\,d\,\left ({\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^4*(a + a/cos(c + d*x))^2),x)

[Out]

-(54*cos(c/2 + (d*x)/2)^2 + 4*cos(c/2 + (d*x)/2)^4 + 24*cos(c/2 + (d*x)/2)^6 - 96*cos(c/2 + (d*x)/2)^8 + 64*co
s(c/2 + (d*x)/2)^10 - 15)/(3360*a^2*d*(cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2) - cos(c/2 + (d*x)/2)^9*sin(c/2
+ (d*x)/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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