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

Optimal. Leaf size=171 \[ \frac {3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {15}{128 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {21 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac {a^2}{40 d (a \sin (c+d x)+a)^5}-\frac {3 a}{64 d (a \sin (c+d x)+a)^4}-\frac {1}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{128 a d (a-a \sin (c+d x))^2}-\frac {5}{64 a d (a \sin (c+d x)+a)^2} \]

[Out]

21/128*arctanh(sin(d*x+c))/a^3/d+1/128/a/d/(a-a*sin(d*x+c))^2-1/40*a^2/d/(a+a*sin(d*x+c))^5-3/64*a/d/(a+a*sin(
d*x+c))^4-1/16/d/(a+a*sin(d*x+c))^3-5/64/a/d/(a+a*sin(d*x+c))^2+3/64/d/(a^3-a^3*sin(d*x+c))-15/128/d/(a^3+a^3*
sin(d*x+c))

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Rubi [A]  time = 0.13, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2667, 44, 206} \[ -\frac {a^2}{40 d (a \sin (c+d x)+a)^5}+\frac {3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {15}{128 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {21 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}-\frac {3 a}{64 d (a \sin (c+d x)+a)^4}-\frac {1}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{128 a d (a-a \sin (c+d x))^2}-\frac {5}{64 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*ArcTanh[Sin[c + d*x]])/(128*a^3*d) + 1/(128*a*d*(a - a*Sin[c + d*x])^2) - a^2/(40*d*(a + a*Sin[c + d*x])^5
) - (3*a)/(64*d*(a + a*Sin[c + d*x])^4) - 1/(16*d*(a + a*Sin[c + d*x])^3) - 5/(64*a*d*(a + a*Sin[c + d*x])^2)
+ 3/(64*d*(a^3 - a^3*Sin[c + d*x])) - 15/(128*d*(a^3 + a^3*Sin[c + d*x]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {1}{64 a^6 (a-x)^3}+\frac {3}{64 a^7 (a-x)^2}+\frac {1}{8 a^3 (a+x)^6}+\frac {3}{16 a^4 (a+x)^5}+\frac {3}{16 a^5 (a+x)^4}+\frac {5}{32 a^6 (a+x)^3}+\frac {15}{128 a^7 (a+x)^2}+\frac {21}{128 a^7 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {1}{128 a d (a-a \sin (c+d x))^2}-\frac {a^2}{40 d (a+a \sin (c+d x))^5}-\frac {3 a}{64 d (a+a \sin (c+d x))^4}-\frac {1}{16 d (a+a \sin (c+d x))^3}-\frac {5}{64 a d (a+a \sin (c+d x))^2}+\frac {3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {15}{128 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {21 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{128 a^2 d}\\ &=\frac {21 \tanh ^{-1}(\sin (c+d x))}{128 a^3 d}+\frac {1}{128 a d (a-a \sin (c+d x))^2}-\frac {a^2}{40 d (a+a \sin (c+d x))^5}-\frac {3 a}{64 d (a+a \sin (c+d x))^4}-\frac {1}{16 d (a+a \sin (c+d x))^3}-\frac {5}{64 a d (a+a \sin (c+d x))^2}+\frac {3}{64 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {15}{128 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 145, normalized size = 0.85 \[ \frac {\sec ^4(c+d x) \left (-105 \sin ^6(c+d x)-315 \sin ^5(c+d x)-140 \sin ^4(c+d x)+420 \sin ^3(c+d x)+469 \sin ^2(c+d x)+7 \sin (c+d x)+105 \tanh ^{-1}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^{10}-176\right )}{640 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*(-176 + 105*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^10 + 7*Sin[c + d*x] + 469*Sin[c + d*x]^2 + 420*Sin[c + d*x]^3 - 140*Sin[c + d*x]^4 - 315*Sin
[c + d*x]^5 - 105*Sin[c + d*x]^6))/(640*a^3*d*(1 + Sin[c + d*x])^3)

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fricas [A]  time = 0.84, size = 248, normalized size = 1.45 \[ -\frac {210 \, \cos \left (d x + c\right )^{6} - 910 \, \cos \left (d x + c\right )^{4} + 252 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} + {\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (3 \, \cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4} + {\left (\cos \left (d x + c\right )^{6} - 4 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 14 \, {\left (45 \, \cos \left (d x + c\right )^{4} - 30 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 96}{1280 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4} + {\left (a^{3} d \cos \left (d x + c\right )^{6} - 4 \, a^{3} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(210*cos(d*x + c)^6 - 910*cos(d*x + c)^4 + 252*cos(d*x + c)^2 - 105*(3*cos(d*x + c)^6 - 4*cos(d*x + c)
^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(sin(d*x + c) + 1) + 105*(3*cos(d*x + c)^6 - 4*cos(d
*x + c)^4 + (cos(d*x + c)^6 - 4*cos(d*x + c)^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 14*(45*cos(d*x + c)^4 -
30*cos(d*x + c)^2 - 16)*sin(d*x + c) + 96)/(3*a^3*d*cos(d*x + c)^6 - 4*a^3*d*cos(d*x + c)^4 + (a^3*d*cos(d*x
+ c)^6 - 4*a^3*d*cos(d*x + c)^4)*sin(d*x + c))

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giac [A]  time = 0.71, size = 136, normalized size = 0.80 \[ \frac {\frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac {10 \, {\left (63 \, \sin \left (d x + c\right )^{2} - 150 \, \sin \left (d x + c\right ) + 91\right )}}{a^{3} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {959 \, \sin \left (d x + c\right )^{5} + 5395 \, \sin \left (d x + c\right )^{4} + 12390 \, \sin \left (d x + c\right )^{3} + 14710 \, \sin \left (d x + c\right )^{2} + 9275 \, \sin \left (d x + c\right ) + 2647}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{5120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5120*(420*log(abs(sin(d*x + c) + 1))/a^3 - 420*log(abs(sin(d*x + c) - 1))/a^3 + 10*(63*sin(d*x + c)^2 - 150*
sin(d*x + c) + 91)/(a^3*(sin(d*x + c) - 1)^2) - (959*sin(d*x + c)^5 + 5395*sin(d*x + c)^4 + 12390*sin(d*x + c)
^3 + 14710*sin(d*x + c)^2 + 9275*sin(d*x + c) + 2647)/(a^3*(sin(d*x + c) + 1)^5))/d

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maple [A]  time = 0.27, size = 162, normalized size = 0.95 \[ \frac {1}{128 a^{3} d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3}{64 a^{3} d \left (\sin \left (d x +c \right )-1\right )}-\frac {21 \ln \left (\sin \left (d x +c \right )-1\right )}{256 a^{3} d}-\frac {1}{40 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {3}{64 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{64 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {15}{128 a^{3} d \left (1+\sin \left (d x +c \right )\right )}+\frac {21 \ln \left (1+\sin \left (d x +c \right )\right )}{256 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128/a^3/d/(sin(d*x+c)-1)^2-3/64/a^3/d/(sin(d*x+c)-1)-21/256/a^3/d*ln(sin(d*x+c)-1)-1/40/a^3/d/(1+sin(d*x+c))
^5-3/64/a^3/d/(1+sin(d*x+c))^4-1/16/a^3/d/(1+sin(d*x+c))^3-5/64/a^3/d/(1+sin(d*x+c))^2-15/128/a^3/d/(1+sin(d*x
+c))+21/256*ln(1+sin(d*x+c))/a^3/d

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maxima [A]  time = 0.60, size = 188, normalized size = 1.10 \[ -\frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{6} + 315 \, \sin \left (d x + c\right )^{5} + 140 \, \sin \left (d x + c\right )^{4} - 420 \, \sin \left (d x + c\right )^{3} - 469 \, \sin \left (d x + c\right )^{2} - 7 \, \sin \left (d x + c\right ) + 176\right )}}{a^{3} \sin \left (d x + c\right )^{7} + 3 \, a^{3} \sin \left (d x + c\right )^{6} + a^{3} \sin \left (d x + c\right )^{5} - 5 \, a^{3} \sin \left (d x + c\right )^{4} - 5 \, a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{1280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1280*(2*(105*sin(d*x + c)^6 + 315*sin(d*x + c)^5 + 140*sin(d*x + c)^4 - 420*sin(d*x + c)^3 - 469*sin(d*x +
c)^2 - 7*sin(d*x + c) + 176)/(a^3*sin(d*x + c)^7 + 3*a^3*sin(d*x + c)^6 + a^3*sin(d*x + c)^5 - 5*a^3*sin(d*x +
c)^4 - 5*a^3*sin(d*x + c)^3 + a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x + c) + a^3) - 105*log(sin(d*x + c) + 1)/a^3
+ 105*log(sin(d*x + c) - 1)/a^3)/d

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mupad [B]  time = 4.77, size = 173, normalized size = 1.01 \[ \frac {21\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{128\,a^3\,d}-\frac {\frac {21\,{\sin \left (c+d\,x\right )}^6}{128}+\frac {63\,{\sin \left (c+d\,x\right )}^5}{128}+\frac {7\,{\sin \left (c+d\,x\right )}^4}{32}-\frac {21\,{\sin \left (c+d\,x\right )}^3}{32}-\frac {469\,{\sin \left (c+d\,x\right )}^2}{640}-\frac {7\,\sin \left (c+d\,x\right )}{640}+\frac {11}{40}}{d\,\left (a^3\,{\sin \left (c+d\,x\right )}^7+3\,a^3\,{\sin \left (c+d\,x\right )}^6+a^3\,{\sin \left (c+d\,x\right )}^5-5\,a^3\,{\sin \left (c+d\,x\right )}^4-5\,a^3\,{\sin \left (c+d\,x\right )}^3+a^3\,{\sin \left (c+d\,x\right )}^2+3\,a^3\,\sin \left (c+d\,x\right )+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a*sin(c + d*x))^3),x)

[Out]

(21*atanh(sin(c + d*x)))/(128*a^3*d) - ((7*sin(c + d*x)^4)/32 - (469*sin(c + d*x)^2)/640 - (21*sin(c + d*x)^3)
/32 - (7*sin(c + d*x))/640 + (63*sin(c + d*x)^5)/128 + (21*sin(c + d*x)^6)/128 + 11/40)/(d*(3*a^3*sin(c + d*x)
+ a^3 + a^3*sin(c + d*x)^2 - 5*a^3*sin(c + d*x)^3 - 5*a^3*sin(c + d*x)^4 + a^3*sin(c + d*x)^5 + 3*a^3*sin(c +
d*x)^6 + a^3*sin(c + d*x)^7))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**5/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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