3.875 \(\int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=77 \[ \frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

[Out]

-a^3*ln(1-sin(d*x+c))/d+a^3*ln(sin(d*x+c))/d+1/2*a^5/d/(a-a*sin(d*x+c))^2+a^4/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 44} \[ \frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

-((a^3*Log[1 - Sin[c + d*x]])/d) + (a^3*Log[Sin[c + d*x]])/d + a^5/(2*d*(a - a*Sin[c + d*x])^2) + a^4/(d*(a -
a*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2836

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 54, normalized size = 0.70 \[ \frac {a^3 \left (\frac {3-2 \sin (c+d x)}{(\sin (c+d x)-1)^2}-2 \log (1-\sin (c+d x))+2 \log (\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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fricas [A]  time = 0.49, size = 126, normalized size = 1.64 \[ \frac {2 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3} + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 123, normalized size = 1.60 \[ -\frac {12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 76 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 114 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 76 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - (25*a^3*tan(1/2*d*x +
1/2*c)^4 - 76*a^3*tan(1/2*d*x + 1/2*c)^3 + 114*a^3*tan(1/2*d*x + 1/2*c)^2 - 76*a^3*tan(1/2*d*x + 1/2*c) + 25*
a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d

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maple [B]  time = 0.61, size = 172, normalized size = 2.23 \[ \frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \sin \left (d x +c \right )}{8 d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3}}{d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {a^{3}}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*a^3*sin(d*x+c)^3/cos(d*x+c)^2+1/8*a^3*sin(d*x+c)/d+1/d*a^3*ln(sec(d*
x+c)+tan(d*x+c))+1/d*a^3/cos(d*x+c)^4+3/4/d*a^3*tan(d*x+c)*sec(d*x+c)^3+9/8/d*a^3*tan(d*x+c)*sec(d*x+c)+1/2/d*
a^3/cos(d*x+c)^2+1/d*a^3*ln(tan(d*x+c))

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maxima [A]  time = 0.48, size = 70, normalized size = 0.91 \[ -\frac {2 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(2*a^3*log(sin(d*x + c) - 1) - 2*a^3*log(sin(d*x + c)) + (2*a^3*sin(d*x + c) - 3*a^3)/(sin(d*x + c)^2 - 2
*sin(d*x + c) + 1))/d

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mupad [B]  time = 9.12, size = 61, normalized size = 0.79 \[ \frac {2\,a^3\,\mathrm {atanh}\left (2\,\sin \left (c+d\,x\right )-1\right )}{d}-\frac {a^3\,\sin \left (c+d\,x\right )-\frac {3\,a^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^3/(cos(c + d*x)^5*sin(c + d*x)),x)

[Out]

(2*a^3*atanh(2*sin(c + d*x) - 1))/d - (a^3*sin(c + d*x) - (3*a^3)/2)/(d*(sin(c + d*x)^2 - 2*sin(c + d*x) + 1))

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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)*sec(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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