3.94 \(\int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=102 \[ -\frac {\cos ^5(c+d x)}{5 a^3 d}+\frac {3 \cos ^4(c+d x)}{4 a^3 d}-\frac {4 \cos ^3(c+d x)}{3 a^3 d}+\frac {2 \cos ^2(c+d x)}{a^3 d}-\frac {4 \cos (c+d x)}{a^3 d}+\frac {4 \log (\cos (c+d x)+1)}{a^3 d} \]

[Out]

-4*cos(d*x+c)/a^3/d+2*cos(d*x+c)^2/a^3/d-4/3*cos(d*x+c)^3/a^3/d+3/4*cos(d*x+c)^4/a^3/d-1/5*cos(d*x+c)^5/a^3/d+
4*ln(1+cos(d*x+c))/a^3/d

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Rubi [A]  time = 0.18, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2836, 12, 88} \[ -\frac {\cos ^5(c+d x)}{5 a^3 d}+\frac {3 \cos ^4(c+d x)}{4 a^3 d}-\frac {4 \cos ^3(c+d x)}{3 a^3 d}+\frac {2 \cos ^2(c+d x)}{a^3 d}-\frac {4 \cos (c+d x)}{a^3 d}+\frac {4 \log (\cos (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(-4*Cos[c + d*x])/(a^3*d) + (2*Cos[c + d*x]^2)/(a^3*d) - (4*Cos[c + d*x]^3)/(3*a^3*d) + (3*Cos[c + d*x]^4)/(4*
a^3*d) - Cos[c + d*x]^5/(5*a^3*d) + (4*Log[1 + Cos[c + d*x]])/(a^3*d)

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]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.96, size = 73, normalized size = 0.72 \[ \frac {-4920 \cos (c+d x)+1320 \cos (2 (c+d x))-380 \cos (3 (c+d x))+90 \cos (4 (c+d x))-12 \cos (5 (c+d x))+7680 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3857}{960 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(3857 - 4920*Cos[c + d*x] + 1320*Cos[2*(c + d*x)] - 380*Cos[3*(c + d*x)] + 90*Cos[4*(c + d*x)] - 12*Cos[5*(c +
d*x)] + 7680*Log[Cos[(c + d*x)/2]])/(960*a^3*d)

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fricas [A]  time = 0.82, size = 70, normalized size = 0.69 \[ -\frac {12 \, \cos \left (d x + c\right )^{5} - 45 \, \cos \left (d x + c\right )^{4} + 80 \, \cos \left (d x + c\right )^{3} - 120 \, \cos \left (d x + c\right )^{2} + 240 \, \cos \left (d x + c\right ) - 240 \, \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{60 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(12*cos(d*x + c)^5 - 45*cos(d*x + c)^4 + 80*cos(d*x + c)^3 - 120*cos(d*x + c)^2 + 240*cos(d*x + c) - 240
*log(1/2*cos(d*x + c) + 1/2))/(a^3*d)

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giac [A]  time = 0.42, size = 172, normalized size = 1.69 \[ -\frac {\frac {60 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{3}} + \frac {\frac {85 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {20 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {200 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {205 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 29}{a^{3} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/15*(60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^3 + (85*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
- 20*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 200*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 205*(cos(d*x
+ c) - 1)^4/(cos(d*x + c) + 1)^4 - 137*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 29)/(a^3*((cos(d*x + c) -
1)/(cos(d*x + c) + 1) - 1)^5))/d

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maple [A]  time = 0.72, size = 114, normalized size = 1.12 \[ -\frac {1}{5 d \,a^{3} \sec \left (d x +c \right )^{5}}+\frac {3}{4 d \,a^{3} \sec \left (d x +c \right )^{4}}-\frac {4}{3 d \,a^{3} \sec \left (d x +c \right )^{3}}+\frac {2}{d \,a^{3} \sec \left (d x +c \right )^{2}}-\frac {4}{d \,a^{3} \sec \left (d x +c \right )}-\frac {4 \ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}}+\frac {4 \ln \left (1+\sec \left (d x +c \right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x)

[Out]

-1/5/d/a^3/sec(d*x+c)^5+3/4/d/a^3/sec(d*x+c)^4-4/3/d/a^3/sec(d*x+c)^3+2/d/a^3/sec(d*x+c)^2-4/d/a^3/sec(d*x+c)-
4/d/a^3*ln(sec(d*x+c))+4/d/a^3*ln(1+sec(d*x+c))

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maxima [A]  time = 0.33, size = 73, normalized size = 0.72 \[ -\frac {\frac {12 \, \cos \left (d x + c\right )^{5} - 45 \, \cos \left (d x + c\right )^{4} + 80 \, \cos \left (d x + c\right )^{3} - 120 \, \cos \left (d x + c\right )^{2} + 240 \, \cos \left (d x + c\right )}{a^{3}} - \frac {240 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*((12*cos(d*x + c)^5 - 45*cos(d*x + c)^4 + 80*cos(d*x + c)^3 - 120*cos(d*x + c)^2 + 240*cos(d*x + c))/a^3
- 240*log(cos(d*x + c) + 1)/a^3)/d

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mupad [B]  time = 0.90, size = 82, normalized size = 0.80 \[ \frac {\frac {4\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\cos \left (c+d\,x\right )}{a^3}+\frac {2\,{\cos \left (c+d\,x\right )}^2}{a^3}-\frac {4\,{\cos \left (c+d\,x\right )}^3}{3\,a^3}+\frac {3\,{\cos \left (c+d\,x\right )}^4}{4\,a^3}-\frac {{\cos \left (c+d\,x\right )}^5}{5\,a^3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^5/(a + a/cos(c + d*x))^3,x)

[Out]

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

[Out]

Timed out

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