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

Optimal. Leaf size=101 \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {3 a^3}{4 d (a-a \sin (c+d x))}-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 101, 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, 72} \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {3 a^3}{4 d (a-a \sin (c+d x))}-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a^2*Log[1 - Sin[c + d*x]])/(8*d) + (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) + (3*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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

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Mathematica [A]  time = 0.29, size = 66, normalized size = 0.65 \[ -\frac {a^2 \left (\frac {6}{\sin (c+d x)-1}-\frac {2}{(\sin (c+d x)-1)^2}+7 \log (1-\sin (c+d x))-8 \log (\sin (c+d x))+\log (\sin (c+d x)+1)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a^2*(7*Log[1 - Sin[c + d*x]] - 8*Log[Sin[c + d*x]] + Log[1 + Sin[c + d*x]] - 2/(-1 + Sin[c + d*x])^2 + 6
/(-1 + Sin[c + d*x])))/d

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fricas [A]  time = 0.48, size = 166, normalized size = 1.64 \[ \frac {6 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2} + 8 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/8*(6*a^2*sin(d*x + c) - 8*a^2 + 8*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(1/2*sin(d*x + c)) -
(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(sin(d*x + c) + 1) - 7*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*
x + c) - 2*a^2)*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 = 0.28, size = 91, normalized size = 0.90 \[ -\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 14 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {21 \, a^{2} \sin \left (d x + c\right )^{2} - 54 \, a^{2} \sin \left (d x + c\right ) + 37 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*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)) + 14*a^2*log(abs(sin(d*x + c) - 1)) - 16*a^2*log(abs(sin(d*x + c))) -
(21*a^2*sin(d*x + c)^2 - 54*a^2*sin(d*x + c) + 37*a^2)/(sin(d*x + c) - 1)^2)/d

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maple [A]  time = 0.66, size = 112, normalized size = 1.11 \[ \frac {a^{2}}{2 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{4 d}+\frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2}}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.48, size = 84, normalized size = 0.83 \[ -\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.08, size = 91, normalized size = 0.90 \[ \frac {a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {\frac {3\,a^2\,\sin \left (c+d\,x\right )}{4}-a^2}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )}-\frac {7\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d} \]

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)),x)

[Out]

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

[Out]

Timed out

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