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

Optimal. Leaf size=110 \[ \frac{3 a^3 \tan (c+d x)}{d}+\frac{11 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{17 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))}-\frac{2 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))^2}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

(11*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (2*a^3*Sin[c + d*x])/(3*d*(1 - Cos[c + d*x])^2) - (17*a^3*Sin[c + d*x])
/(3*d*(1 - Cos[c + d*x])) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.229634, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2872, 2650, 2648, 3770, 3767, 8, 3768} \[ \frac{3 a^3 \tan (c+d x)}{d}+\frac{11 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{17 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))}-\frac{2 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))^2}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (2*a^3*Sin[c + d*x])/(3*d*(1 - Cos[c + d*x])^2) - (17*a^3*Sin[c + d*x])
/(3*d*(1 - Cos[c + d*x])) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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 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 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 3770

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

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 8

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rubi steps

\begin{align*} \int \csc ^4(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^4(c+d x) \sec ^3(c+d x) \, dx\\ &=a^4 \int \left (\frac{2}{a (1-\cos (c+d x))^2}+\frac{5}{a (1-\cos (c+d x))}+\frac{5 \sec (c+d x)}{a}+\frac{3 \sec ^2(c+d x)}{a}+\frac{\sec ^3(c+d x)}{a}\right ) \, dx\\ &=a^3 \int \sec ^3(c+d x) \, dx+\left (2 a^3\right ) \int \frac{1}{(1-\cos (c+d x))^2} \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx+\left (5 a^3\right ) \int \frac{1}{1-\cos (c+d x)} \, dx+\left (5 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac{5 a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))^2}-\frac{5 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} a^3 \int \sec (c+d x) \, dx+\frac{1}{3} \left (2 a^3\right ) \int \frac{1}{1-\cos (c+d x)} \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac{11 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{2 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))^2}-\frac{17 a^3 \sin (c+d x)}{3 d (1-\cos (c+d x))}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 6.23361, size = 678, normalized size = 6.16 \[ \frac{3 \sin \left (\frac{d x}{2}\right ) \cos ^3(c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{8 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{3 \sin \left (\frac{d x}{2}\right ) \cos ^3(c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{8 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\cos ^3(c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{32 d \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}-\frac{\cos ^3(c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{32 d \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}-\frac{11 \cos ^3(c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3 \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{16 d}+\frac{11 \cos ^3(c+d x) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3 \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{16 d}-\frac{\cot \left (\frac{c}{2}\right ) \cos ^3(c+d x) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{24 d}+\frac{\csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^3(c+d x) \csc ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{24 d}+\frac{17 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^3(c+d x) \csc \left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^3}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[c + d*x]^3*Cot[c/2]*Csc[c/2 + (d*x)/2]^2*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/(24*d) - (11*Cos[c
+ d*x]^3*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/(16*d) + (
11*Cos[c + d*x]^3*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/(1
6*d) + (17*Cos[c + d*x]^3*Csc[c/2]*Csc[c/2 + (d*x)/2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*Sin[(d*x)/2]
)/(24*d) + (Cos[c + d*x]^3*Csc[c/2]*Csc[c/2 + (d*x)/2]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*Sin[(d*x)
/2])/(24*d) + (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3)/(32*d*(Cos[c/2 + (d*x)/2] - Sin[c/2
+ (d*x)/2])^2) + (3*Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*Sin[(d*x)/2])/(8*d*(Cos[c/2] -
Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) - (Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*
x])^3)/(32*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (3*Cos[c + d*x]^3*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[
c + d*x])^3*Sin[(d*x)/2])/(8*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.074, size = 188, normalized size = 1.7 \begin{align*} -{\frac{26\,{a}^{3}\cot \left ( dx+c \right ) }{3\,d}}-{\frac{{a}^{3}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{{a}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{11\,{a}^{3}}{2\,d\sin \left ( dx+c \right ) }}+{\frac{11\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{{a}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+4\,{\frac{{a}^{3}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{{a}^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,{a}^{3}}{6\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*(a+a*sec(d*x+c))^3,x)

[Out]

-26/3*a^3*cot(d*x+c)/d-1/3/d*a^3*cot(d*x+c)*csc(d*x+c)^2-1/d*a^3/sin(d*x+c)^3-11/2/d*a^3/sin(d*x+c)+11/2/d*a^3
*ln(sec(d*x+c)+tan(d*x+c))-1/d*a^3/sin(d*x+c)^3/cos(d*x+c)+4/d*a^3/sin(d*x+c)/cos(d*x+c)-1/3/d*a^3/sin(d*x+c)^
3/cos(d*x+c)^2+5/6/d*a^3/sin(d*x+c)/cos(d*x+c)^2

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Maxima [A]  time = 1.02512, size = 254, normalized size = 2.31 \begin{align*} -\frac{a^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{3}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3}{\left (\frac{6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac{4 \,{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{3}}{\tan \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/12*(a^3*(2*(15*sin(d*x + c)^4 - 10*sin(d*x + c)^2 - 2)/(sin(d*x + c)^5 - sin(d*x + c)^3) - 15*log(sin(d*x +
c) + 1) + 15*log(sin(d*x + c) - 1)) + 6*a^3*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1
) + 3*log(sin(d*x + c) - 1)) + 12*a^3*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)) + 4*(3*tan(d*x
+ c)^2 + 1)*a^3/tan(d*x + c)^3)/d

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Fricas [A]  time = 1.73043, size = 444, normalized size = 4.04 \begin{align*} -\frac{104 \, a^{3} \cos \left (d x + c\right )^{4} - 38 \, a^{3} \cos \left (d x + c\right )^{3} - 118 \, a^{3} \cos \left (d x + c\right )^{2} + 30 \, a^{3} \cos \left (d x + c\right ) + 6 \, a^{3} - 33 \,{\left (a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 33 \,{\left (a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(104*a^3*cos(d*x + c)^4 - 38*a^3*cos(d*x + c)^3 - 118*a^3*cos(d*x + c)^2 + 30*a^3*cos(d*x + c) + 6*a^3 -
33*(a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(sin(d*x + c) + 1)*sin(d*x + c) + 33*(a^3*cos(d*x + c)^3 - a^
3*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)*sin(d*x + c))/((d*cos(d*x + c)^3 - d*cos(d*x + c)^2)*sin(d*x + c))

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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)**4*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.3604, size = 166, normalized size = 1.51 \begin{align*} \frac{33 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 33 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{6 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac{2 \,{\left (18 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(33*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 33*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(5*a^3*tan(1/2*
d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - 2*(18*a^3*tan(1/2*d*x + 1/2*c)^2
+ a^3)/tan(1/2*d*x + 1/2*c)^3)/d