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

Optimal. Leaf size=126 \[ \frac{\tan ^5(c+d x)}{5 a d}+\frac{\tan ^3(c+d x)}{a d}+\frac{3 \tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d}-\frac{\sec ^5(c+d x)}{5 a d}-\frac{\sec ^3(c+d x)}{3 a d}-\frac{\sec (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.166008, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2839, 2620, 270, 2622, 302, 207} \[ \frac{\tan ^5(c+d x)}{5 a d}+\frac{\tan ^3(c+d x)}{a d}+\frac{3 \tan (c+d x)}{a d}-\frac{\cot (c+d x)}{a d}-\frac{\sec ^5(c+d x)}{5 a d}-\frac{\sec ^3(c+d x)}{3 a d}-\frac{\sec (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(a*d) - Cot[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) + (3*Tan[c + d*x])/(a*d) + Tan[c + d*x]^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 2620

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

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 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])

Rubi steps

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

Mathematica [B]  time = 0.596397, size = 341, normalized size = 2.71 \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (352 \sin (c+d x)-596 \sin (2 (c+d x))+864 \sin (3 (c+d x))-298 \sin (4 (c+d x))+384 \sin (5 (c+d x))+1216 \cos (2 (c+d x))+149 \cos (3 (c+d x))+528 \cos (4 (c+d x))+149 \cos (5 (c+d x))+480 \sin (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+240 \sin (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+120 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+120 \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-120 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-120 \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (240 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-240 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-298\right )-480 \sin (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-240 \sin (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+176\right )}{3840 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(176 + 1216*Cos[2*(c + d*x)] + 149*Cos[3*(c + d*x)] + 528*C
os[4*(c + d*x)] + 149*Cos[5*(c + d*x)] + 120*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 120*Cos[5*(c + d*x)]*Log
[Cos[(c + d*x)/2]] - 120*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 120*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] +
Cos[c + d*x]*(-298 - 240*Log[Cos[(c + d*x)/2]] + 240*Log[Sin[(c + d*x)/2]]) + 352*Sin[c + d*x] - 596*Sin[2*(c
+ d*x)] - 480*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] + 480*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 864*Sin[3
*(c + d*x)] - 298*Sin[4*(c + d*x)] - 240*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] + 240*Log[Sin[(c + d*x)/2]]*Si
n[4*(c + d*x)] + 384*Sin[5*(c + d*x)]))/(3840*a*d*(1 + Sin[c + d*x]))

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Maple [A]  time = 0.102, size = 223, normalized size = 1.8 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\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{9}{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}}-{\frac{7}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{5}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{39}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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)^2*sec(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

1/2/d/a*tan(1/2*d*x+1/2*c)-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)^2-9/8/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)^4-7/3/d/a/(tan(1/2*d*x+1/2*c)+1)^3
+5/2/d/a/(tan(1/2*d*x+1/2*c)+1)^2-39/8/a/d/(tan(1/2*d*x+1/2*c)+1)-1/2/d/a/tan(1/2*d*x+1/2*c)-1/d/a*ln(tan(1/2*
d*x+1/2*c))

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Maxima [B]  time = 1.10039, size = 512, normalized size = 4.06 \begin{align*} -\frac{\frac{\frac{122 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{26 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{454 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{252 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{510 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{330 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{210 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{195 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 15}{\frac{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{2 \, 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 )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{6 \, 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{2 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{15 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((122*sin(d*x + c)/(cos(d*x + c) + 1) - 26*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 454*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 - 252*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 510*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 330*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6 - 210*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 195*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 + 15)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*a*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 - 6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*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 - 2*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - a*sin(d*x + c)^9/(cos(d*x +
c) + 1)^9) + 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 15*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.57092, size = 514, normalized size = 4.08 \begin{align*} \frac{66 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} + 15 \,{\left (\cos \left (d x + c\right )^{5} - \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 )^{5} - \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 (48 \, \cos \left (d x + c\right )^{4} - 9 \, \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 )^{5} - 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)^2*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(66*cos(d*x + c)^4 - 28*cos(d*x + c)^2 + 15*(cos(d*x + c)^5 - 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)^5 - cos(d*x + c)^3*sin(d*x + c) - cos(d*x + c)^3)*log(-1/2*c
os(d*x + c) + 1/2) + 2*(48*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - 1)*sin(d*x + c) - 8)/(a*d*cos(d*x + c)^5 - 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)**2*sec(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.28615, size = 240, normalized size = 1.9 \begin{align*} -\frac{\frac{120 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{60 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{60 \,{\left (2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{5 \,{\left (27 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 25\right )}}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{585 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2040 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2890 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1880 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 493}{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)^2*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 - 60*tan(1/2*d*x + 1/2*c)/a - 60*(2*tan(1/2*d*x + 1/2*c) - 1)/(a*
tan(1/2*d*x + 1/2*c)) + 5*(27*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x + 1/2*c) + 25)/(a*(tan(1/2*d*x + 1/2*c)
- 1)^3) + (585*tan(1/2*d*x + 1/2*c)^4 + 2040*tan(1/2*d*x + 1/2*c)^3 + 2890*tan(1/2*d*x + 1/2*c)^2 + 1880*tan(1
/2*d*x + 1/2*c) + 493)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d