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

Optimal. Leaf size=115 \[ -\frac {\tan ^5(c+d x)}{5 a d}-\frac {2 \tan ^3(c+d x)}{3 a d}-\frac {\tan (c+d x)}{a d}+\frac {\sec ^5(c+d x)}{5 a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec (c+d x)}{a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

[Out]

-arctanh(cos(d*x+c))/a/d+sec(d*x+c)/a/d+1/3*sec(d*x+c)^3/a/d+1/5*sec(d*x+c)^5/a/d-tan(d*x+c)/a/d-2/3*tan(d*x+c
)^3/a/d-1/5*tan(d*x+c)^5/a/d

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Rubi [A]  time = 0.12, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2839, 2622, 302, 207, 3767} \[ -\frac {\tan ^5(c+d x)}{5 a d}-\frac {2 \tan ^3(c+d x)}{3 a d}-\frac {\tan (c+d x)}{a d}+\frac {\sec ^5(c+d x)}{5 a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec (c+d x)}{a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a*d)) + Sec[c + d*x]/(a*d) + Sec[c + d*x]^3/(3*a*d) + Sec[c + d*x]^5/(5*a*d) - Tan[c
+ d*x]/(a*d) - (2*Tan[c + d*x]^3)/(3*a*d) - Tan[c + d*x]^5/(5*a*d)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 302

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 2622

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

Rule 2839

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 3767

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]

Rubi steps

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

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Mathematica [B]  time = 0.66, size = 267, normalized size = 2.32 \[ -\frac {\sec ^3(c+d x) \left (-22 \sin (c+d x)+\frac {149}{4} \sin (2 (c+d x))-14 \sin (3 (c+d x))+\frac {149}{8} \sin (4 (c+d x))-76 \cos (2 (c+d x))+\frac {149}{4} \cos (3 (c+d x))-8 \cos (4 (c+d x))-30 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (-90 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+90 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {447}{4}\right )-30 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-100\right )}{120 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*(Sec[c + d*x]^3*(-100 - 76*Cos[2*(c + d*x)] + (149*Cos[3*(c + d*x)])/4 - 8*Cos[4*(c + d*x)] + 30*Cos[3*
(c + d*x)]*Log[Cos[(c + d*x)/2]] + Cos[c + d*x]*(447/4 + 90*Log[Cos[(c + d*x)/2]] - 90*Log[Sin[(c + d*x)/2]])
- 30*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 22*Sin[c + d*x] + (149*Sin[2*(c + d*x)])/4 + 30*Log[Cos[(c + d*x
)/2]]*Sin[2*(c + d*x)] - 30*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] - 14*Sin[3*(c + d*x)] + (149*Sin[4*(c + d*x
)])/8 + 15*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)]))/(a*d*(1 + Sin[
c + d*x]))

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fricas [A]  time = 0.46, size = 149, normalized size = 1.30 \[ \frac {16 \, \cos \left (d x + c\right )^{4} + 22 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (7 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 8}{30 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(16*cos(d*x + c)^4 + 22*cos(d*x + c)^2 - 15*(cos(d*x + c)^3*sin(d*x + c) + cos(d*x + c)^3)*log(1/2*cos(d*
x + c) + 1/2) + 15*(cos(d*x + c)^3*sin(d*x + c) + cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) + 2*(7*cos(d*x
+ c)^2 + 1)*sin(d*x + c) + 8)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)

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giac [A]  time = 0.21, size = 136, normalized size = 1.18 \[ \frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {5 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {3 \, {\left (115 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 380 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 530 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 340 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 91\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - 5*(21*tan(1/2*d*x + 1/2*c)^2 - 36*tan(1/2*d*x + 1/2*c) + 19)/(a*
(tan(1/2*d*x + 1/2*c) - 1)^3) + 3*(115*tan(1/2*d*x + 1/2*c)^4 + 380*tan(1/2*d*x + 1/2*c)^3 + 530*tan(1/2*d*x +
1/2*c)^2 + 340*tan(1/2*d*x + 1/2*c) + 91)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

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maple [A]  time = 0.50, size = 187, normalized size = 1.63 \[ -\frac {1}{6 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {7}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2}{5 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {23}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/a/d/(tan(1/2*d*x+1/2*c)-1)^3-1/4/a/d/(tan(1/2*d*x+1/2*c)-1)^2-7/8/a/d/(tan(1/2*d*x+1/2*c)-1)+1/a/d*ln(tan
(1/2*d*x+1/2*c))+2/5/a/d/(tan(1/2*d*x+1/2*c)+1)^5-1/a/d/(tan(1/2*d*x+1/2*c)+1)^4+2/a/d/(tan(1/2*d*x+1/2*c)+1)^
3-2/a/d/(tan(1/2*d*x+1/2*c)+1)^2+23/8/a/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 320, normalized size = 2.78 \[ \frac {\frac {2 \, {\left (\frac {31 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {31 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {73 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {65 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 23\right )}}{a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {15 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*(31*sin(d*x + c)/(cos(d*x + c) + 1) - 31*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 73*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 25*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 65*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 23)/(a + 2*a*sin(d*x + c)/(cos(d*x + c
) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a
*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 15*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 11.53, size = 143, normalized size = 1.24 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {146\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}-\frac {62\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {62\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {46}{15}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(c/2 + (d*x)/2))/(a*d) - ((62*tan(c/2 + (d*x)/2))/15 - (62*tan(c/2 + (d*x)/2)^2)/15 - (146*tan(c/2 + (d
*x)/2)^3)/15 + (10*tan(c/2 + (d*x)/2)^4)/3 + (26*tan(c/2 + (d*x)/2)^5)/3 + 2*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2
+ (d*x)/2)^7 + 46/15)/(a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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