3.860 \(\int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=162 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {3 a \csc (c+d x)}{d}-\frac {59 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {11 a \log (\sin (c+d x)+1)}{16 d} \]

[Out]

-3*a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-1/3*a*csc(d*x+c)^3/d-59/16*a*ln(1-sin(d*x+c))/d+3*a*ln(sin(d*x+c))/d+11
/16*a*ln(1+sin(d*x+c))/d+1/8*a^3/d/(a-a*sin(d*x+c))^2+5/4*a^2/d/(a-a*sin(d*x+c))-1/8*a^2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 162, 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, 88} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {3 a \csc (c+d x)}{d}-\frac {59 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {11 a \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(-3*a*Csc[c + d*x])/d - (a*Csc[c + d*x]^2)/(2*d) - (a*Csc[c + d*x]^3)/(3*d) - (59*a*Log[1 - Sin[c + d*x]])/(16
*d) + (3*a*Log[Sin[c + d*x]])/d + (11*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + (5*
a^2)/(4*d*(a - a*Sin[c + d*x])) - a^2/(8*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 ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {a^4}{(a-x)^3 x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \left (\frac {1}{4 a^6 (a-x)^3}+\frac {5}{4 a^7 (a-x)^2}+\frac {59}{16 a^8 (a-x)}+\frac {1}{a^5 x^4}+\frac {1}{a^6 x^3}+\frac {3}{a^7 x^2}+\frac {3}{a^8 x}+\frac {1}{8 a^7 (a+x)^2}+\frac {11}{16 a^8 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {3 a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {59 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {11 a \log (1+\sin (c+d x))}{16 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 1.16, size = 90, normalized size = 0.56 \[ -\frac {a \csc ^3(c+d x) \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\sin ^2(c+d x)\right )}{3 d}-\frac {a \left (2 \csc ^2(c+d x)-\sec ^4(c+d x)-4 \sec ^2(c+d x)-12 \log (\sin (c+d x))+12 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

-1/3*(a*Csc[c + d*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[c + d*x]^2])/d - (a*(2*Csc[c + d*x]^2 + 12*Log[Cos
[c + d*x]] - 12*Log[Sin[c + d*x]] - 4*Sec[c + d*x]^2 - Sec[c + d*x]^4))/(4*d)

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fricas [B]  time = 0.47, size = 343, normalized size = 2.12 \[ -\frac {138 \, a \cos \left (d x + c\right )^{4} - 172 \, a \cos \left (d x + c\right )^{2} - 144 \, {\left (a \cos \left (d x + c\right )^{6} - 2 \, a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} + {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 33 \, {\left (a \cos \left (d x + c\right )^{6} - 2 \, a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} + {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 177 \, {\left (a \cos \left (d x + c\right )^{6} - 2 \, a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} + {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (105 \, a \cos \left (d x + c\right )^{4} - 104 \, a \cos \left (d x + c\right )^{2} + 3 \, a\right )} \sin \left (d x + c\right ) + 18 \, a}{48 \, {\left (d \cos \left (d x + c\right )^{6} - 2 \, d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2} + {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(138*a*cos(d*x + c)^4 - 172*a*cos(d*x + c)^2 - 144*(a*cos(d*x + c)^6 - 2*a*cos(d*x + c)^4 + a*cos(d*x +
c)^2 + (a*cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) - 33*(a*cos(d*x + c)^6 - 2*a*
cos(d*x + c)^4 + a*cos(d*x + c)^2 + (a*cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1)
+ 177*(a*cos(d*x + c)^6 - 2*a*cos(d*x + c)^4 + a*cos(d*x + c)^2 + (a*cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d*
x + c))*log(-sin(d*x + c) + 1) - 2*(105*a*cos(d*x + c)^4 - 104*a*cos(d*x + c)^2 + 3*a)*sin(d*x + c) + 18*a)/(d
*cos(d*x + c)^6 - 2*d*cos(d*x + c)^4 + d*cos(d*x + c)^2 + (d*cos(d*x + c)^4 - d*cos(d*x + c)^2)*sin(d*x + c))

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giac [A]  time = 0.29, size = 149, normalized size = 0.92 \[ \frac {66 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 354 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 288 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {6 \, {\left (11 \, a \sin \left (d x + c\right ) + 13 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {3 \, {\left (177 \, a \sin \left (d x + c\right )^{2} - 394 \, a \sin \left (d x + c\right ) + 221 \, a\right )}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {16 \, {\left (33 \, a \sin \left (d x + c\right )^{3} + 18 \, a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) + 2 \, a\right )}}{\sin \left (d x + c\right )^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(66*a*log(abs(sin(d*x + c) + 1)) - 354*a*log(abs(sin(d*x + c) - 1)) + 288*a*log(abs(sin(d*x + c))) - 6*(1
1*a*sin(d*x + c) + 13*a)/(sin(d*x + c) + 1) + 3*(177*a*sin(d*x + c)^2 - 394*a*sin(d*x + c) + 221*a)/(sin(d*x +
c) - 1)^2 - 16*(33*a*sin(d*x + c)^3 + 18*a*sin(d*x + c)^2 + 3*a*sin(d*x + c) + 2*a)/sin(d*x + c)^3)/d

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maple [A]  time = 0.40, size = 173, normalized size = 1.07 \[ \frac {a}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3 a}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3 a}{2 d \sin \left (d x +c \right )^{2}}+\frac {3 a \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a}{4 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7 a}{12 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {35 a}{24 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {35 a}{8 d \sin \left (d x +c \right )}+\frac {35 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

1/4/d*a/sin(d*x+c)^2/cos(d*x+c)^4+3/4/d*a/sin(d*x+c)^2/cos(d*x+c)^2-3/2/d*a/sin(d*x+c)^2+3*a*ln(tan(d*x+c))/d+
1/4/d*a/sin(d*x+c)^3/cos(d*x+c)^4-7/12/d*a/sin(d*x+c)^3/cos(d*x+c)^2+35/24/d*a/sin(d*x+c)/cos(d*x+c)^2-35/8*a/
d/sin(d*x+c)+35/8/d*a*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.37, size = 138, normalized size = 0.85 \[ \frac {33 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 177 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, a \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (105 \, a \sin \left (d x + c\right )^{5} - 69 \, a \sin \left (d x + c\right )^{4} - 106 \, a \sin \left (d x + c\right )^{3} + 52 \, a \sin \left (d x + c\right )^{2} + 4 \, a \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{6} - \sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(33*a*log(sin(d*x + c) + 1) - 177*a*log(sin(d*x + c) - 1) + 144*a*log(sin(d*x + c)) - 2*(105*a*sin(d*x +
c)^5 - 69*a*sin(d*x + c)^4 - 106*a*sin(d*x + c)^3 + 52*a*sin(d*x + c)^2 + 4*a*sin(d*x + c) + 8*a)/(sin(d*x + c
)^6 - sin(d*x + c)^5 - sin(d*x + c)^4 + sin(d*x + c)^3))/d

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mupad [B]  time = 0.11, size = 145, normalized size = 0.90 \[ \frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {35\,a\,{\sin \left (c+d\,x\right )}^5}{8}-\frac {23\,a\,{\sin \left (c+d\,x\right )}^4}{8}-\frac {53\,a\,{\sin \left (c+d\,x\right )}^3}{12}+\frac {13\,a\,{\sin \left (c+d\,x\right )}^2}{6}+\frac {a\,\sin \left (c+d\,x\right )}{6}+\frac {a}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^6-{\sin \left (c+d\,x\right )}^5-{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^3\right )}-\frac {59\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}+\frac {11\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)^4),x)

[Out]

(3*a*log(sin(c + d*x)))/d - (a/3 + (a*sin(c + d*x))/6 + (13*a*sin(c + d*x)^2)/6 - (53*a*sin(c + d*x)^3)/12 - (
23*a*sin(c + d*x)^4)/8 + (35*a*sin(c + d*x)^5)/8)/(d*(sin(c + d*x)^3 - sin(c + d*x)^4 - sin(c + d*x)^5 + sin(c
+ d*x)^6)) - (59*a*log(sin(c + d*x) - 1))/(16*d) + (11*a*log(sin(c + d*x) + 1))/(16*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)**4*sec(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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