3.3.2 \(\int \frac {\csc ^5(c+d x)}{a+b \sec (c+d x)} \, dx\) [202]

Optimal. Leaf size=179 \[ \frac {\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac {a (3 a+b) \log (1-\cos (c+d x))}{16 (a+b)^3 d}-\frac {a (3 a-b) \log (1+\cos (c+d x))}{16 (a-b)^3 d}+\frac {a^4 b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d} \]

[Out]

1/8*(4*a^2*b-a*(3*a^2+b^2)*cos(d*x+c))*csc(d*x+c)^2/(a^2-b^2)^2/d+1/4*(b-a*cos(d*x+c))*csc(d*x+c)^4/(a^2-b^2)/
d+1/16*a*(3*a+b)*ln(1-cos(d*x+c))/(a+b)^3/d-1/16*a*(3*a-b)*ln(1+cos(d*x+c))/(a-b)^3/d+a^4*b*ln(b+a*cos(d*x+c))
/(a^2-b^2)^3/d

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Rubi [A]
time = 0.22, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2916, 12, 837, 815} \begin {gather*} \frac {\csc ^4(c+d x) (b-a \cos (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\csc ^2(c+d x) \left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a^4 b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac {a (3 a+b) \log (1-\cos (c+d x))}{16 d (a+b)^3}-\frac {a (3 a-b) \log (\cos (c+d x)+1)}{16 d (a-b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2*b - a*(3*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(8*(a^2 - b^2)^2*d) + ((b - a*Cos[c + d*x])*Csc[c +
d*x]^4)/(4*(a^2 - b^2)*d) + (a*(3*a + b)*Log[1 - Cos[c + d*x]])/(16*(a + b)^3*d) - (a*(3*a - b)*Log[1 + Cos[c
+ d*x]])/(16*(a - b)^3*d) + (a^4*b*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[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 ^5(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cot (c+d x) \csc ^4(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac {a^5 \text {Subst}\left (\int \frac {x}{a (-b+x) \left (a^2-x^2\right )^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^4 \text {Subst}\left (\int \frac {x}{(-b+x) \left (a^2-x^2\right )^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac {a^2 \text {Subst}\left (\int \frac {a^2 b+3 a^2 x}{(-b+x) \left (a^2-x^2\right )^2} \, dx,x,-a \cos (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=\frac {\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \frac {a^2 b \left (5 a^2-b^2\right )+a^2 \left (3 a^2+b^2\right ) x}{(-b+x) \left (a^2-x^2\right )} \, dx,x,-a \cos (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=\frac {\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac {\text {Subst}\left (\int \left (\frac {a (3 a-b) (a+b)^2}{2 (a-b) (a-x)}-\frac {8 a^4 b}{(a-b) (a+b) (b-x)}+\frac {a (a-b)^2 (3 a+b)}{2 (a+b) (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=\frac {\left (4 a^2 b-a \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{8 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^4(c+d x)}{4 \left (a^2-b^2\right ) d}+\frac {a (3 a+b) \log (1-\cos (c+d x))}{16 (a+b)^3 d}-\frac {a (3 a-b) \log (1+\cos (c+d x))}{16 (a-b)^3 d}+\frac {a^4 b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^3 d}\\ \end {align*}

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Mathematica [A]
time = 3.06, size = 207, normalized size = 1.16 \begin {gather*} \frac {-2 (a-b)^3 \left (3 a^2+4 a b+b^2\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )-(a-b)^3 (a+b)^2 \csc ^4\left (\frac {1}{2} (c+d x)\right )+8 a \left (-\left ((3 a-b) (a+b)^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 a^3 b \log (b+a \cos (c+d x))+(a-b)^3 (3 a+b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 (a+b)^3 \left (3 a^2-4 a b+b^2\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+(a-b)^2 (a+b)^3 \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 (a-b)^3 (a+b)^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b)^3*(3*a^2 + 4*a*b + b^2)*Csc[(c + d*x)/2]^2 - (a - b)^3*(a + b)^2*Csc[(c + d*x)/2]^4 + 8*a*(-((3*a
- b)*(a + b)^3*Log[Cos[(c + d*x)/2]]) + 8*a^3*b*Log[b + a*Cos[c + d*x]] + (a - b)^3*(3*a + b)*Log[Sin[(c + d*x
)/2]]) + 2*(a + b)^3*(3*a^2 - 4*a*b + b^2)*Sec[(c + d*x)/2]^2 + (a - b)^2*(a + b)^3*Sec[(c + d*x)/2]^4)/(64*(a
- b)^3*(a + b)^3*d)

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Maple [A]
time = 0.19, size = 172, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {-3 a +b}{16 \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {\left (3 a -b \right ) a \ln \left (1+\cos \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {1}{2 \left (8 a +8 b \right ) \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {-3 a -b}{16 \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (3 a +b \right ) a \ln \left (-1+\cos \left (d x +c \right )\right )}{16 \left (a +b \right )^{3}}+\frac {b \,a^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(172\)
default	\(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {-3 a +b}{16 \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {\left (3 a -b \right ) a \ln \left (1+\cos \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {1}{2 \left (8 a +8 b \right ) \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {-3 a -b}{16 \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (3 a +b \right ) a \ln \left (-1+\cos \left (d x +c \right )\right )}{16 \left (a +b \right )^{3}}+\frac {b \,a^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(172\)
norman	\(\frac {-\frac {1}{64 d \left (a +b \right )}+\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d \left (a -b \right )}+\frac {\left (2 a -b \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \left (a^{2}-2 b a +b^{2}\right )}-\frac {\left (2 a +b \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \left (a^{2}+2 b a +b^{2}\right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {a^{4} b \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {\left (3 a +b \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}\)	\(230\)
risch	\(\frac {3 i a^{2} c}{8 d \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}-\frac {i a b x}{8 \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}-\frac {2 i a^{4} b x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {i a b x}{8 \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}-\frac {3 i a^{2} x}{8 \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}-\frac {2 i a^{4} b c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {i a b c}{8 d \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}+\frac {3 i a^{2} x}{8 \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}-\frac {3 i a^{2} c}{8 d \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}-\frac {i a b c}{8 d \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}-\frac {-3 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-b^{2} a \,{\mathrm e}^{7 i \left (d x +c \right )}+8 b \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+11 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-7 b^{2} a \,{\mathrm e}^{5 i \left (d x +c \right )}-32 b \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+11 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-7 b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}+8 b \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{3} {\mathrm e}^{i \left (d x +c \right )}-b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}}{4 \left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{8 d \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{8 d \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}+\frac {a^{4} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(754\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/2/(8*a-8*b)/(1+cos(d*x+c))^2-1/16*(-3*a+b)/(a-b)^2/(1+cos(d*x+c))-1/16*(3*a-b)*a/(a-b)^3*ln(1+cos(d*x+c
))-1/2/(8*a+8*b)/(-1+cos(d*x+c))^2-1/16*(-3*a-b)/(a+b)^2/(-1+cos(d*x+c))+1/16*(3*a+b)/(a+b)^3*a*ln(-1+cos(d*x+
c))+b*a^4/(a+b)^3/(a-b)^3*ln(b+a*cos(d*x+c)))

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Maxima [A]
time = 0.27, size = 268, normalized size = 1.50 \begin {gather*} \frac {\frac {16 \, a^{4} b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a^{2} - a b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + a b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (4 \, a^{2} b \cos \left (d x + c\right )^{2} - {\left (3 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} - 6 \, a^{2} b + 2 \, b^{3} + {\left (5 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^4*b*log(a*cos(d*x + c) + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (3*a^2 - a*b)*log(cos(d*x + c) +
1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + a*b)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(
4*a^2*b*cos(d*x + c)^2 - (3*a^3 + a*b^2)*cos(d*x + c)^3 - 6*a^2*b + 2*b^3 + (5*a^3 - a*b^2)*cos(d*x + c))/((a^
4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (172) = 344\).
time = 3.25, size = 469, normalized size = 2.62 \begin {gather*} \frac {12 \, a^{4} b - 16 \, a^{2} b^{3} + 4 \, b^{5} + 2 \, {\left (3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} - 8 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) + 16 \, {\left (a^{4} b \cos \left (d x + c\right )^{4} - 2 \, a^{4} b \cos \left (d x + c\right )^{2} + a^{4} b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4} + {\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4} + {\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(12*a^4*b - 16*a^2*b^3 + 4*b^5 + 2*(3*a^5 - 2*a^3*b^2 - a*b^4)*cos(d*x + c)^3 - 8*(a^4*b - a^2*b^3)*cos(d
*x + c)^2 - 2*(5*a^5 - 6*a^3*b^2 + a*b^4)*cos(d*x + c) + 16*(a^4*b*cos(d*x + c)^4 - 2*a^4*b*cos(d*x + c)^2 + a
^4*b)*log(a*cos(d*x + c) + b) - (3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4 + (3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*c
os(d*x + c)^4 - 2*(3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (3*a^5 -
8*a^4*b + 6*a^3*b^2 - a*b^4 + (3*a^5 - 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4 - 2*(3*a^5 - 8*a^4*b + 6*a
^3*b^2 - a*b^4)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x +
c)^4 - 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{5}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(c + d*x)**5/(a + b*sec(c + d*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (172) = 344\).
time = 0.48, size = 419, normalized size = 2.34 \begin {gather*} \frac {\frac {64 \, a^{4} b \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {4 \, {\left (3 \, a^{2} + a b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {\frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a^{2} + 2 \, a b + b^{2} - \frac {8 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {12 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {18 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(64*a^4*b*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) +
1)))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 4*(3*a^2 + a*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/
(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(a^2 -
2*a*b + b^2) - (a^2 + 2*a*b + b^2 - 8*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*a*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - 4*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 18*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) +
1)^2 + 6*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*
(cos(d*x + c) - 1)^2))/d

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Mupad [B]
time = 1.61, size = 297, normalized size = 1.66 \begin {gather*} \frac {\frac {3\,a^2\,b-b^3}{4\,{\left (a^2-b^2\right )}^2}+\frac {{\cos \left (c+d\,x\right )}^3\,\left (3\,a^3+a\,b^2\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {a^2\,b\,{\cos \left (c+d\,x\right )}^2}{2\,{\left (a^2-b^2\right )}^2}-\frac {a\,\cos \left (c+d\,x\right )\,\left (5\,a^2-b^2\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^4-2\,{\cos \left (c+d\,x\right )}^2+1\right )}+\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (\frac {3}{16\,\left (a+b\right )}-\frac {5\,b}{16\,{\left (a+b\right )}^2}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}+\frac {5\,b}{16\,{\left (a-b\right )}^2}+\frac {3}{16\,\left (a-b\right )}\right )}{d}+\frac {a^4\,b\,\ln \left (b+a\,\cos \left (c+d\,x\right )\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^5*(a + b/cos(c + d*x))),x)

[Out]

((3*a^2*b - b^3)/(4*(a^2 - b^2)^2) + (cos(c + d*x)^3*(a*b^2 + 3*a^3))/(8*(a^4 + b^4 - 2*a^2*b^2)) - (a^2*b*cos
(c + d*x)^2)/(2*(a^2 - b^2)^2) - (a*cos(c + d*x)*(5*a^2 - b^2))/(8*(a^4 + b^4 - 2*a^2*b^2)))/(d*(cos(c + d*x)^
4 - 2*cos(c + d*x)^2 + 1)) + (log(cos(c + d*x) - 1)*(3/(16*(a + b)) - (5*b)/(16*(a + b)^2) + b^2/(8*(a + b)^3)
))/d - (log(cos(c + d*x) + 1)*(b^2/(8*(a - b)^3) + (5*b)/(16*(a - b)^2) + 3/(16*(a - b))))/d + (a^4*b*log(b +
a*cos(c + d*x)))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))

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