3.816 \(\int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=86 \[ -\frac {a^3 \cot (c+d x)}{d}+\frac {11 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

-3*a^3*arctanh(cos(d*x+c))/d-a^3*cot(d*x+c)/d+2/3*a^3*cos(d*x+c)/d/(1-sin(d*x+c))^2+11/3*a^3*cos(d*x+c)/d/(1-s
in(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {2872, 3770, 3767, 8, 2650, 2648} \[ -\frac {a^3 \cot (c+d x)}{d}+\frac {11 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*a^3*ArcTanh[Cos[c + d*x]])/d - (a^3*Cot[c + d*x])/d + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])^2) + (1
1*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/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]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \csc ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=a^4 \int \left (\frac {3 \csc (c+d x)}{a}+\frac {\csc ^2(c+d x)}{a}+\frac {2}{a (-1+\sin (c+d x))^2}-\frac {3}{a (-1+\sin (c+d x))}\right ) \, dx\\ &=a^3 \int \csc ^2(c+d x) \, dx+\left (2 a^3\right ) \int \frac {1}{(-1+\sin (c+d x))^2} \, dx+\left (3 a^3\right ) \int \csc (c+d x) \, dx-\left (3 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx\\ &=-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {3 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {1}{3} \left (2 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\frac {a^3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {11 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.96, size = 135, normalized size = 1.57 \[ \frac {a^3 \left (3 \tan \left (\frac {1}{2} (c+d x)\right )-3 \cot \left (\frac {1}{2} (c+d x)\right )+18 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right ) (11 \sin (c+d x)-13)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(-3*Cot[(c + d*x)/2] - 18*Log[Cos[(c + d*x)/2]] + 18*Log[Sin[(c + d*x)/2]] + 4/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^2 + (4*Sin[(c + d*x)/2]*(-13 + 11*Sin[c + d*x]))/(-Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 3*Tan[
(c + d*x)/2]))/(6*d)

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 334, normalized size = 3.88 \[ \frac {28 \, a^{3} \cos \left (d x + c\right )^{3} - 10 \, a^{3} \cos \left (d x + c\right )^{2} - 34 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3} - 9 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + 2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + 2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (14 \, a^{3} \cos \left (d x + c\right )^{2} + 19 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/6*(28*a^3*cos(d*x + c)^3 - 10*a^3*cos(d*x + c)^2 - 34*a^3*cos(d*x + c) + 4*a^3 - 9*(a^3*cos(d*x + c)^3 + 2*a
^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3 - (a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3)*sin(d*x + c))*l
og(1/2*cos(d*x + c) + 1/2) + 9*(a^3*cos(d*x + c)^3 + 2*a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3 - (a^3*co
s(d*x + c)^2 - a^3*cos(d*x + c) - 2*a^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(14*a^3*cos(d*x + c)^2
+ 19*a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 - d*cos(d*x + c) - (d*cos
(d*x + c)^2 - d*cos(d*x + c) - 2*d)*sin(d*x + c) - 2*d)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 118, normalized size = 1.37 \[ \frac {18 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {4 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/6*(18*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^3*tan(1/2*d*x + 1/2*c) - 3*(6*a^3*tan(1/2*d*x + 1/2*c) + a^3)
/tan(1/2*d*x + 1/2*c) - 4*(15*a^3*tan(1/2*d*x + 1/2*c)^2 - 24*a^3*tan(1/2*d*x + 1/2*c) + 13*a^3)/(tan(1/2*d*x
+ 1/2*c) - 1)^3)/d

________________________________________________________________________________________

maple [A]  time = 0.72, size = 155, normalized size = 1.80 \[ \frac {4 a^{3}}{3 d \cos \left (d x +c \right )^{3}}+\frac {2 a^{3} \tan \left (d x +c \right )}{d}+\frac {a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 a^{3}}{d \cos \left (d x +c \right )}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a^{3}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4 a^{3}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 a^{3} \cot \left (d x +c \right )}{3 d} \]

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))^3,x)

[Out]

4/3/d*a^3/cos(d*x+c)^3+2*a^3*tan(d*x+c)/d+1/d*a^3*tan(d*x+c)*sec(d*x+c)^2+3/d*a^3/cos(d*x+c)+3/d*a^3*ln(csc(d*
x+c)-cot(d*x+c))+1/3/d*a^3/sin(d*x+c)/cos(d*x+c)^3+4/3/d*a^3/sin(d*x+c)/cos(d*x+c)-8/3*a^3*cot(d*x+c)/d

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 123, normalized size = 1.43 \[ \frac {2 \, {\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{3} + 6 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \, a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, a^{3}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 9.21, size = 144, normalized size = 1.67 \[ \frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-21\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {61\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+a^3}{d\,\left (-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^3/(cos(c + d*x)^4*sin(c + d*x)^2),x)

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d - (35*a^3*tan(c/2 + (d*x)/2)^2 - 21*a^3*tan(c/2 + (d*x)/2)^3 + a^3 - (61*a^3
*tan(c/2 + (d*x)/2))/3)/(d*(2*tan(c/2 + (d*x)/2) - 6*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2
+ (d*x)/2)^4)) + (a^3*tan(c/2 + (d*x)/2))/(2*d)

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________