3.802 \(\int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=81 \[ \frac {a \tan ^3(c+d x)}{3 d}+\frac {2 a \tan (c+d x)}{d}-\frac {a \cot (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x)}{d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2838, 2620, 270, 2622, 302, 207} \[ \frac {a \tan ^3(c+d x)}{3 d}+\frac {2 a \tan (c+d x)}{d}-\frac {a \cot (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x)}{d}-\frac {a \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTanh[Cos[c + d*x]])/d) - (a*Cot[c + d*x])/d + (a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) + (2*a*Tan
[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*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 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 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 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 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 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 109, normalized size = 1.35 \[ \frac {5 a \tan (c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}+\frac {a \sec (c+d x)}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cot[c + d*x])/d) - (a*Log[Cos[(c + d*x)/2]])/d + (a*Log[Sin[(c + d*x)/2]])/d + (a*Sec[c + d*x])/d + (a*Se
c[c + d*x]^3)/(3*d) + (5*a*Tan[c + d*x])/(3*d) + (a*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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fricas [B]  time = 0.48, size = 170, normalized size = 2.10 \[ -\frac {10 \, a \cos \left (d x + c\right )^{2} + 3 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (8 \, a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - 4 \, a}{6 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

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]

-1/6*(10*a*cos(d*x + c)^2 + 3*(a*cos(d*x + c)^3 + a*cos(d*x + c)*sin(d*x + c) - a*cos(d*x + c))*log(1/2*cos(d*
x + c) + 1/2) - 3*(a*cos(d*x + c)^3 + a*cos(d*x + c)*sin(d*x + c) - a*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/
2) - 2*(8*a*cos(d*x + c)^2 - a)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^3 + d*cos(d*x + c)*sin(d*x + c) - d*cos(d*
x + c))

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giac [A]  time = 0.21, size = 129, normalized size = 1.59 \[ \frac {6 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {21 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

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]

1/6*(6*a*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a*tan(1/2*d*x + 1/2*c) - 3*(a*tan(1/2*d*x + 1/2*c)^2 + 3*a*tan(1/2
*d*x + 1/2*c) + a)/(tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c)) - (21*a*tan(1/2*d*x + 1/2*c)^2 - 36*a*tan(1
/2*d*x + 1/2*c) + 19*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [A]  time = 0.46, size = 106, normalized size = 1.31 \[ \frac {a}{3 d \cos \left (d x +c \right )^{3}}+\frac {a}{d \cos \left (d x +c \right )}+\frac {a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4 a}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 a \cot \left (d x +c \right )}{3 d} \]

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]

1/3/d*a/cos(d*x+c)^3+1/d*a/cos(d*x+c)+1/d*a*ln(csc(d*x+c)-cot(d*x+c))+1/3/d*a/sin(d*x+c)/cos(d*x+c)^3+4/3/d*a/
sin(d*x+c)/cos(d*x+c)-8/3*a*cot(d*x+c)/d

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maxima [A]  time = 0.32, size = 83, normalized size = 1.02 \[ \frac {2 \, {\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a + a {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

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]

1/6*(2*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a + a*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*l
og(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 9.10, size = 145, normalized size = 1.79 \[ \frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {22\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+a}{d\,\left (-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))/(cos(c + d*x)^4*sin(c + d*x)^2),x)

[Out]

(a*tan(c/2 + (d*x)/2))/(2*d) + (a*log(tan(c/2 + (d*x)/2)))/d - (a - (22*a*tan(c/2 + (d*x)/2))/3 + (8*a*tan(c/2
+ (d*x)/2)^2)/3 + 10*a*tan(c/2 + (d*x)/2)^3 - 9*a*tan(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2) - 4*tan(c/2
+ (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^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)**2*sec(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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