3.1.70 \(\int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx\) [70]

Optimal. Leaf size=55 \[ \frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot ^5(c+d x)}{5 a d}-\frac {\csc ^5(c+d x)}{5 a d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3957, 2918, 2686, 30, 2687, 14} \begin {gather*} \frac {\cot ^5(c+d x)}{5 a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {\csc ^5(c+d x)}{5 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Cot[c + d*x]^3/(3*a*d) + Cot[c + d*x]^5/(5*a*d) - Csc[c + d*x]^5/(5*a*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

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 2687

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 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^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*} \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx &=-\int \frac {\cot (c+d x) \csc ^3(c+d x)}{-a-a \cos (c+d x)} \, dx\\ &=-\frac {\int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a}+\frac {\int \cot (c+d x) \csc ^5(c+d x) \, dx}{a}\\ &=-\frac {\text {Subst}\left (\int x^4 \, dx,x,\csc (c+d x)\right )}{a d}-\frac {\text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=-\frac {\csc ^5(c+d x)}{5 a d}-\frac {\text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot ^5(c+d x)}{5 a d}-\frac {\csc ^5(c+d x)}{5 a d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(55)=110\).
time = 0.35, size = 116, normalized size = 2.11 \begin {gather*} -\frac {\csc (c) \csc ^3(c+d x) \sec (c+d x) (240 \sin (c)-96 \sin (d x)-54 \sin (c+d x)-18 \sin (2 (c+d x))+18 \sin (3 (c+d x))+9 \sin (4 (c+d x))-32 \sin (c+2 d x)+32 \sin (2 c+3 d x)+16 \sin (3 c+4 d x))}{960 a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/960*(Csc[c]*Csc[c + d*x]^3*Sec[c + d*x]*(240*Sin[c] - 96*Sin[d*x] - 54*Sin[c + d*x] - 18*Sin[2*(c + d*x)] +
18*Sin[3*(c + d*x)] + 9*Sin[4*(c + d*x)] - 32*Sin[c + 2*d*x] + 32*Sin[2*c + 3*d*x] + 16*Sin[3*c + 4*d*x]))/(a
*d*(1 + Sec[c + d*x]))

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Maple [A]
time = 0.10, size = 62, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{16 d a}\)	\(62\)
default	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{16 d a}\)	\(62\)
norman	\(\frac {-\frac {1}{48 a d}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}-\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{80 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}\)	\(79\)
risch	\(\frac {4 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}-1\right )}{15 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{3}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d/a*(-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)

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Maxima [A]
time = 0.27, size = 96, normalized size = 1.75 \begin {gather*} -\frac {\frac {\frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} + \frac {5 \, {\left (\frac {6 \, \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 \sin \left (d x + c\right )^{3}}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*((10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a + 5*(6*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d

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Fricas [A]
time = 3.21, size = 89, normalized size = 1.62 \begin {gather*} -\frac {2 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 3}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )} \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(2*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 3*cos(d*x + c)^2 - 3*cos(d*x + c) - 3)/((a*d*cos(d*x + c)^3 + a*d
*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)*sin(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.45, size = 74, normalized size = 1.35 \begin {gather*} -\frac {\frac {5 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{a^{5}}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(5*(6*tan(1/2*d*x + 1/2*c)^2 + 1)/(a*tan(1/2*d*x + 1/2*c)^3) + (3*a^4*tan(1/2*d*x + 1/2*c)^5 + 10*a^4*t
an(1/2*d*x + 1/2*c)^3)/a^5)/d

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Mupad [B]
time = 1.09, size = 60, normalized size = 1.09 \begin {gather*} -\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+5}{240\,a\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(30*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^6 + 3*tan(c/2 + (d*x)/2)^8 + 5)/(240*a*d*tan(c/2 + (d*x)/2)^
3)

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