3.868 \(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=134 \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {7 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}-\frac {31 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {4 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[Out]

-2*a^2*csc(d*x+c)/d-1/2*a^2*csc(d*x+c)^2/d-31/8*a^2*ln(1-sin(d*x+c))/d+4*a^2*ln(sin(d*x+c))/d-1/8*a^2*ln(1+sin
(d*x+c))/d+1/4*a^4/d/(a-a*sin(d*x+c))^2+7/4*a^3/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {7 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}-\frac {31 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {4 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(-2*a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) - (31*a^2*Log[1 - Sin[c + d*x]])/(8*d) + (4*a^2*Log[Sin[c
+ d*x]])/d - (a^2*Log[1 + Sin[c + d*x]])/(8*d) + a^4/(4*d*(a - a*Sin[c + d*x])^2) + (7*a^3)/(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {a^3}{(a-x)^3 x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^8 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 x^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^8 \operatorname {Subst}\left (\int \left (\frac {1}{2 a^4 (a-x)^3}+\frac {7}{4 a^5 (a-x)^2}+\frac {31}{8 a^6 (a-x)}+\frac {1}{a^4 x^3}+\frac {2}{a^5 x^2}+\frac {4}{a^6 x}-\frac {1}{8 a^6 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {31 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {4 a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {7 a^3}{4 d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 84, normalized size = 0.63 \[ -\frac {a^2 \left (\frac {14}{\sin (c+d x)-1}-\frac {2}{(\sin (c+d x)-1)^2}+4 \csc ^2(c+d x)+16 \csc (c+d x)+31 \log (1-\sin (c+d x))-32 \log (\sin (c+d x))+\log (\sin (c+d x)+1)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

-1/8*(a^2*(16*Csc[c + d*x] + 4*Csc[c + d*x]^2 + 31*Log[1 - Sin[c + d*x]] - 32*Log[Sin[c + d*x]] + Log[1 + Sin[
c + d*x]] - 2/(-1 + Sin[c + d*x])^2 + 14/(-1 + Sin[c + d*x])))/d

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fricas [B]  time = 0.47, size = 302, normalized size = 2.25 \[ -\frac {44 \, a^{2} \cos \left (d x + c\right )^{2} - 40 \, a^{2} - 32 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 31 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 19 \, a^{2}\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{4} - 3 \, d \cos \left (d x + c\right )^{2} + 2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) + 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/8*(44*a^2*cos(d*x + c)^2 - 40*a^2 - 32*(a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 + 2*a^2 + 2*(a^2*cos(d*x
+ c)^2 - a^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) + (a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 + 2*a^2 + 2*(a^
2*cos(d*x + c)^2 - a^2)*sin(d*x + c))*log(sin(d*x + c) + 1) + 31*(a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 +
2*a^2 + 2*(a^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 2*(15*a^2*cos(d*x + c)^2 - 19*a^2)
*sin(d*x + c))/(d*cos(d*x + c)^4 - 3*d*cos(d*x + c)^2 + 2*(d*cos(d*x + c)^2 - d)*sin(d*x + c) + 2*d)

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giac [A]  time = 0.30, size = 125, normalized size = 0.93 \[ -\frac {4 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 124 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 128 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4} + 114 \, a^{2} \sin \left (d x + c\right )^{3} - 173 \, a^{2} \sin \left (d x + c\right )^{2} + 32 \, a^{2} \sin \left (d x + c\right ) + 16 \, a^{2}}{{\left (\sin \left (d x + c\right )^{2} - \sin \left (d x + c\right )\right )}^{2}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/32*(4*a^2*log(abs(sin(d*x + c) + 1)) + 124*a^2*log(abs(sin(d*x + c) - 1)) - 128*a^2*log(abs(sin(d*x + c)))
+ (3*a^2*sin(d*x + c)^4 + 114*a^2*sin(d*x + c)^3 - 173*a^2*sin(d*x + c)^2 + 32*a^2*sin(d*x + c) + 16*a^2)/(sin
(d*x + c)^2 - sin(d*x + c))^2)/d

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maple [A]  time = 0.57, size = 199, normalized size = 1.49 \[ \frac {a^{2}}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{2}}{2 d \cos \left (d x +c \right )^{2}}+\frac {4 a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a^{2}}{2 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 a^{2}}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 a^{2}}{4 d \sin \left (d x +c \right )}+\frac {15 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2}}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3 a^{2}}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3 a^{2}}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x)

[Out]

1/4/d*a^2/cos(d*x+c)^4+1/2/d*a^2/cos(d*x+c)^2+4/d*a^2*ln(tan(d*x+c))+1/2/d*a^2/sin(d*x+c)/cos(d*x+c)^4+5/4/d*a
^2/sin(d*x+c)/cos(d*x+c)^2-15/4/d*a^2/sin(d*x+c)+15/4/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*a^2/sin(d*x+c)^2/c
os(d*x+c)^4+3/4/d*a^2/sin(d*x+c)^2/cos(d*x+c)^2-3/2/d*a^2/sin(d*x+c)^2

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maxima [A]  time = 0.31, size = 119, normalized size = 0.89 \[ -\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 31 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - 32 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, a^{2} \sin \left (d x + c\right )^{3} - 22 \, a^{2} \sin \left (d x + c\right )^{2} + 4 \, a^{2} \sin \left (d x + c\right ) + 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8*(a^2*log(sin(d*x + c) + 1) + 31*a^2*log(sin(d*x + c) - 1) - 32*a^2*log(sin(d*x + c)) + 2*(15*a^2*sin(d*x
+ c)^3 - 22*a^2*sin(d*x + c)^2 + 4*a^2*sin(d*x + c) + 2*a^2)/(sin(d*x + c)^4 - 2*sin(d*x + c)^3 + sin(d*x + c)
^2))/d

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mupad [B]  time = 0.10, size = 126, normalized size = 0.94 \[ \frac {4\,a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {\frac {15\,a^2\,{\sin \left (c+d\,x\right )}^3}{4}-\frac {11\,a^2\,{\sin \left (c+d\,x\right )}^2}{2}+a^2\,\sin \left (c+d\,x\right )+\frac {a^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2\right )}-\frac {31\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^5*sin(c + d*x)^3),x)

[Out]

(4*a^2*log(sin(c + d*x)))/d - (a^2*log(sin(c + d*x) + 1))/(8*d) - (a^2*sin(c + d*x) + a^2/2 - (11*a^2*sin(c +
d*x)^2)/2 + (15*a^2*sin(c + d*x)^3)/4)/(d*(sin(c + d*x)^2 - 2*sin(c + d*x)^3 + sin(c + d*x)^4)) - (31*a^2*log(
sin(c + d*x) - 1))/(8*d)

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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)**3*sec(d*x+c)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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