3.869 \(\int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=150 \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {9 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {4 a^2 \csc (c+d x)}{d}-\frac {49 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {6 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[Out]

-4*a^2*csc(d*x+c)/d-a^2*csc(d*x+c)^2/d-1/3*a^2*csc(d*x+c)^3/d-49/8*a^2*ln(1-sin(d*x+c))/d+6*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+9/4*a^3/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 150, 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} \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {9 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {4 a^2 \csc (c+d x)}{d}-\frac {49 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {6 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(-4*a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/d - (a^2*Csc[c + d*x]^3)/(3*d) - (49*a^2*Log[1 - Sin[c + d*x]])
/(8*d) + (6*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) +
(9*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 ^4(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^4}{(a-x)^3 x^4 (a+x)} \, 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)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \left (\frac {1}{2 a^5 (a-x)^3}+\frac {9}{4 a^6 (a-x)^2}+\frac {49}{8 a^7 (a-x)}+\frac {1}{a^4 x^4}+\frac {2}{a^5 x^3}+\frac {4}{a^6 x^2}+\frac {6}{a^7 x}+\frac {1}{8 a^7 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {4 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {49 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {6 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 {9 a^3}{4 d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.06, size = 133, normalized size = 0.89 \[ \frac {a^9 \left (-\frac {\csc ^3(c+d x)}{3 a^7}-\frac {\csc ^2(c+d x)}{a^7}-\frac {4 \csc (c+d x)}{a^7}-\frac {49 \log (1-\sin (c+d x))}{8 a^7}+\frac {6 \log (\sin (c+d x))}{a^7}+\frac {\log (\sin (c+d x)+1)}{8 a^7}+\frac {9}{4 a^6 (a-a \sin (c+d x))}+\frac {1}{4 a^5 (a-a \sin (c+d x))^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^9*((-4*Csc[c + d*x])/a^7 - Csc[c + d*x]^2/a^7 - Csc[c + d*x]^3/(3*a^7) - (49*Log[1 - Sin[c + d*x]])/(8*a^7)
+ (6*Log[Sin[c + d*x]])/a^7 + Log[1 + Sin[c + d*x]]/(8*a^7) + 1/(4*a^5*(a - a*Sin[c + d*x])^2) + 9/(4*a^6*(a
- a*Sin[c + d*x]))))/d

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fricas [B]  time = 0.50, size = 370, normalized size = 2.47 \[ \frac {150 \, a^{2} \cos \left (d x + c\right )^{4} - 356 \, a^{2} \cos \left (d x + c\right )^{2} + 214 \, a^{2} + 144 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 3 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 147 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (57 \, a^{2} \cos \left (d x + c\right )^{2} - 55 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, {\left (2 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - 3 \, d \cos \left (d x + c\right )^{2} + 2 \, d\right )} \sin \left (d x + c\right ) + 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(150*a^2*cos(d*x + c)^4 - 356*a^2*cos(d*x + c)^2 + 214*a^2 + 144*(2*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x +
c)^2 + 2*a^2 - (a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 + 2*a^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) + 3*(2*
a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 + 2*a^2 - (a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 + 2*a^2)*sin(d*
x + c))*log(sin(d*x + c) + 1) - 147*(2*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 + 2*a^2 - (a^2*cos(d*x + c)^4
- 3*a^2*cos(d*x + c)^2 + 2*a^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 4*(57*a^2*cos(d*x + c)^2 - 55*a^2)*sin
(d*x + c))/(2*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - 3*d*cos(d*x + c)^2 + 2*d)*sin(d*x +
c) + 2*d)

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giac [A]  time = 0.35, size = 142, normalized size = 0.95 \[ \frac {6 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 294 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 288 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {3 \, {\left (147 \, a^{2} \sin \left (d x + c\right )^{2} - 330 \, a^{2} \sin \left (d x + c\right ) + 187 \, a^{2}\right )}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {16 \, {\left (33 \, a^{2} \sin \left (d x + c\right )^{3} + 12 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{3}}}{48 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/48*(6*a^2*log(abs(sin(d*x + c) + 1)) - 294*a^2*log(abs(sin(d*x + c) - 1)) + 288*a^2*log(abs(sin(d*x + c))) +
3*(147*a^2*sin(d*x + c)^2 - 330*a^2*sin(d*x + c) + 187*a^2)/(sin(d*x + c) - 1)^2 - 16*(33*a^2*sin(d*x + c)^3
+ 12*a^2*sin(d*x + c)^2 + 3*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^3)/d

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maple [A]  time = 0.56, size = 215, normalized size = 1.43 \[ \frac {a^{2}}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {25 a^{2}}{12 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {25 a^{2}}{4 d \sin \left (d x +c \right )}+\frac {25 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3 a^{2}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3 a^{2}}{d \sin \left (d x +c \right )^{2}}+\frac {6 a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a^{2}}{4 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7 a^{2}}{12 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}} \]

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))^2,x)

[Out]

1/4/d*a^2/sin(d*x+c)/cos(d*x+c)^4+25/12/d*a^2/sin(d*x+c)/cos(d*x+c)^2-25/4/d*a^2/sin(d*x+c)+25/4/d*a^2*ln(sec(
d*x+c)+tan(d*x+c))+1/2/d*a^2/sin(d*x+c)^2/cos(d*x+c)^4+3/2/d*a^2/sin(d*x+c)^2/cos(d*x+c)^2-3/d*a^2/sin(d*x+c)^
2+6/d*a^2*ln(tan(d*x+c))+1/4/d*a^2/sin(d*x+c)^3/cos(d*x+c)^4-7/12/d*a^2/sin(d*x+c)^3/cos(d*x+c)^2

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maxima [A]  time = 0.45, size = 133, normalized size = 0.89 \[ \frac {3 \, a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 147 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (75 \, a^{2} \sin \left (d x + c\right )^{4} - 114 \, a^{2} \sin \left (d x + c\right )^{3} + 28 \, a^{2} \sin \left (d x + c\right )^{2} + 4 \, a^{2} \sin \left (d x + c\right ) + 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{3}}}{24 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

1/24*(3*a^2*log(sin(d*x + c) + 1) - 147*a^2*log(sin(d*x + c) - 1) + 144*a^2*log(sin(d*x + c)) - 2*(75*a^2*sin(
d*x + c)^4 - 114*a^2*sin(d*x + c)^3 + 28*a^2*sin(d*x + c)^2 + 4*a^2*sin(d*x + c) + 4*a^2)/(sin(d*x + c)^5 - 2*
sin(d*x + c)^4 + sin(d*x + c)^3))/d

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mupad [B]  time = 9.05, size = 140, normalized size = 0.93 \[ \frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {49\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d}-\frac {\frac {25\,a^2\,{\sin \left (c+d\,x\right )}^4}{4}-\frac {19\,a^2\,{\sin \left (c+d\,x\right )}^3}{2}+\frac {7\,a^2\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{3}+\frac {a^2}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^3\right )}+\frac {6\,a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \]

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)^4),x)

[Out]

(a^2*log(sin(c + d*x) + 1))/(8*d) - (49*a^2*log(sin(c + d*x) - 1))/(8*d) - ((a^2*sin(c + d*x))/3 + a^2/3 + (7*
a^2*sin(c + d*x)^2)/3 - (19*a^2*sin(c + d*x)^3)/2 + (25*a^2*sin(c + d*x)^4)/4)/(d*(sin(c + d*x)^3 - 2*sin(c +
d*x)^4 + sin(c + d*x)^5)) + (6*a^2*log(sin(c + d*x)))/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))**2,x)

[Out]

Timed out

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