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

Optimal. Leaf size=116 \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {5 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc (c+d x)}{d}-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[Out]

-a^2*csc(d*x+c)/d-17/8*a^2*ln(1-sin(d*x+c))/d+2*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+5/4*a^3/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 116, 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} \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {5 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc (c+d x)}{d}-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Csc[c + d*x])/d) - (17*a^2*Log[1 - Sin[c + d*x]])/(8*d) + (2*a^2*Log[Sin[c + d*x]])/d + (a^2*Log[1 + Si
n[c + d*x]])/(8*d) + a^4/(4*d*(a - a*Sin[c + d*x])^2) + (5*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 ^2(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^2}{(a-x)^3 x^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 x^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {1}{2 a^3 (a-x)^3}+\frac {5}{4 a^4 (a-x)^2}+\frac {17}{8 a^5 (a-x)}+\frac {1}{a^4 x^2}+\frac {2}{a^5 x}+\frac {1}{8 a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {a^2 \csc (c+d x)}{d}-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {2 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 {5 a^3}{4 d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 74, normalized size = 0.64 \[ \frac {a^2 \left (-\frac {10}{\sin (c+d x)-1}+\frac {2}{(\sin (c+d x)-1)^2}-8 \csc (c+d x)-17 \log (1-\sin (c+d x))+16 \log (\sin (c+d x))+\log (\sin (c+d x)+1)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-8*Csc[c + d*x] - 17*Log[1 - Sin[c + d*x]] + 16*Log[Sin[c + d*x]] + Log[1 + Sin[c + d*x]] + 2/(-1 + Sin[
c + d*x])^2 - 10/(-1 + Sin[c + d*x])))/(8*d)

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fricas [B]  time = 0.51, size = 240, normalized size = 2.07 \[ \frac {18 \, a^{2} \cos \left (d x + c\right )^{2} + 28 \, a^{2} \sin \left (d x + c\right ) - 26 \, a^{2} + 16 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 17 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (2 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(18*a^2*cos(d*x + c)^2 + 28*a^2*sin(d*x + c) - 26*a^2 + 16*(2*a^2*cos(d*x + c)^2 - 2*a^2 - (a^2*cos(d*x +
c)^2 - 2*a^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) + (2*a^2*cos(d*x + c)^2 - 2*a^2 - (a^2*cos(d*x + c)^2 - 2*a^
2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 17*(2*a^2*cos(d*x + c)^2 - 2*a^2 - (a^2*cos(d*x + c)^2 - 2*a^2)*sin(d
*x + c))*log(-sin(d*x + c) + 1))/(2*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 2*d)*sin(d*x + c) - 2*d)

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giac [A]  time = 0.30, size = 115, normalized size = 0.99 \[ \frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 34 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 32 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {16 \, {\left (2 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )} + \frac {51 \, a^{2} \sin \left (d x + c\right )^{2} - 122 \, a^{2} \sin \left (d x + c\right ) + 75 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) - 34*a^2*log(abs(sin(d*x + c) - 1)) + 32*a^2*log(abs(sin(d*x + c))) - 1
6*(2*a^2*sin(d*x + c) + a^2)/sin(d*x + c) + (51*a^2*sin(d*x + c)^2 - 122*a^2*sin(d*x + c) + 75*a^2)/(sin(d*x +
c) - 1)^2)/d

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maple [A]  time = 0.62, size = 176, normalized size = 1.52 \[ \frac {a^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {9 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2}}{2 d \cos \left (d x +c \right )^{4}}+\frac {a^{2}}{d \cos \left (d x +c \right )^{2}}+\frac {2 a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a^{2}}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 a^{2}}{8 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 a^{2}}{8 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*a^2*tan(d*x+c)*sec(d*x+c)^3+3/8/d*a^2*tan(d*x+c)*sec(d*x+c)+9/4/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^
2/cos(d*x+c)^4+1/d*a^2/cos(d*x+c)^2+2/d*a^2*ln(tan(d*x+c))+1/4/d*a^2/sin(d*x+c)/cos(d*x+c)^4+5/8/d*a^2/sin(d*x
+c)/cos(d*x+c)^2-15/8/d*a^2/sin(d*x+c)

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maxima [A]  time = 0.31, size = 104, normalized size = 0.90 \[ \frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 17 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (9 \, a^{2} \sin \left (d x + c\right )^{2} - 14 \, a^{2} \sin \left (d x + c\right ) + 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{3} - 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*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) - 17*a^2*log(sin(d*x + c) - 1) + 16*a^2*log(sin(d*x + c)) - 2*(9*a^2*sin(d*x +
c)^2 - 14*a^2*sin(d*x + c) + 4*a^2)/(sin(d*x + c)^3 - 2*sin(d*x + c)^2 + sin(d*x + c)))/d

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mupad [B]  time = 9.05, size = 110, normalized size = 0.95 \[ \frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {17\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d}+\frac {2\,a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {9\,a^2\,{\sin \left (c+d\,x\right )}^2}{4}-\frac {7\,a^2\,\sin \left (c+d\,x\right )}{2}+a^2}{d\,\left ({\sin \left (c+d\,x\right )}^3-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\right )} \]

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)^2),x)

[Out]

(a^2*log(sin(c + d*x) + 1))/(8*d) - (17*a^2*log(sin(c + d*x) - 1))/(8*d) + (2*a^2*log(sin(c + d*x)))/d - (a^2
- (7*a^2*sin(c + d*x))/2 + (9*a^2*sin(c + d*x)^2)/4)/(d*(sin(c + d*x) - 2*sin(c + d*x)^2 + sin(c + d*x)^3))

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

[Out]

Timed out

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