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[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]))

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Rubi [A]  time = 0.164349, antiderivative size = 150, normalized size of antiderivative = 1., 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 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]

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]))

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*}

Mathematica [A]  time = 6.05746, size = 133, normalized size = 0.89 \[ \frac{a^9 \left (\frac{9}{4 a^6 (a-a \sin (c+d x))}+\frac{1}{4 a^5 (a-a \sin (c+d x))^2}-\frac{\csc ^3(c+d x)}{3 a^7}-\frac{\csc ^2(c+d x)}{a^7}-\frac{4 \csc (c+d x)}{a^7}-\frac{49 \log (1-\sin (c+d x))}{8 a^7}+\frac{6 \log (\sin (c+d x))}{a^7}+\frac{\log (\sin (c+d x)+1)}{8 a^7}\right )}{d} \]

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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Maple [A]  time = 0.152, size = 215, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{25\,{a}^{2}}{12\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{25\,{a}^{2}}{4\,d\sin \left ( dx+c \right ) }}+{\frac{25\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{{a}^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+6\,{\frac{{a}^{2}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{7\,{a}^{2}}{12\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

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 = 1.13435, size = 180, normalized size = 1.2 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 1.65177, size = 906, normalized size = 6.04 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.25799, size = 192, normalized size = 1.28 \begin{align*} \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} \end{align*}

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