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[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)

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Rubi [A]  time = 0.124173, antiderivative size = 115, normalized size of antiderivative = 1., 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 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 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 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 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 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*}

Mathematica [B]  time = 0.639078, 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]

-(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*C
os[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)]))/(120*a*d*(1 + Sin[c
+ d*x]))

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Maple [A]  time = 0.098, size = 187, normalized size = 1.6 \begin{align*} -{\frac{1}{6\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{4\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{7}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{2}{5\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+2\,{\frac{1}{da \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-2\,{\frac{1}{da \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+{\frac{23}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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

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Maxima [B]  time = 1.03177, size = 432, normalized size = 3.76 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.39562, size = 413, normalized size = 3.59 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.20221, size = 184, normalized size = 1.6 \begin{align*} \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} \end{align*}

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