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

Optimal. Leaf size=120 \[ -\frac {a^2}{24 d (a \sin (c+d x)+a)^3}+\frac {a}{32 d (a-a \sin (c+d x))^2}-\frac {3 a}{32 d (a \sin (c+d x)+a)^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a \sin (c+d x)+a)}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d} \]

[Out]

5/16*arctanh(sin(d*x+c))/a/d+1/32*a/d/(a-a*sin(d*x+c))^2+1/8/d/(a-a*sin(d*x+c))-1/24*a^2/d/(a+a*sin(d*x+c))^3-
3/32*a/d/(a+a*sin(d*x+c))^2-3/16/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 120, 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} \[ -\frac {a^2}{24 d (a \sin (c+d x)+a)^3}+\frac {a}{32 d (a-a \sin (c+d x))^2}-\frac {3 a}{32 d (a \sin (c+d x)+a)^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a \sin (c+d x)+a)}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sin[c + d*x]])/(16*a*d) + a/(32*d*(a - a*Sin[c + d*x])^2) + 1/(8*d*(a - a*Sin[c + d*x])) - a^2/(24*
d*(a + a*Sin[c + d*x])^3) - (3*a)/(32*d*(a + a*Sin[c + d*x])^2) - 3/(16*d*(a + a*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)} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {1}{16 a^4 (a-x)^3}+\frac {1}{8 a^5 (a-x)^2}+\frac {1}{8 a^3 (a+x)^4}+\frac {3}{16 a^4 (a+x)^3}+\frac {3}{16 a^5 (a+x)^2}+\frac {5}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 97, normalized size = 0.81 \[ \frac {\sec ^4(c+d x) \left (-15 \sin ^4(c+d x)-15 \sin ^3(c+d x)+25 \sin ^2(c+d x)+25 \sin (c+d x)+15 (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^3 \tanh ^{-1}(\sin (c+d x))-8\right )}{48 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*(-8 + 25*Sin[c + d*x] + 25*Sin[c + d*x]^2 - 15*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 15*ArcTanh
[Sin[c + d*x]]*(-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^3))/(48*a*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.64, size = 147, normalized size = 1.22 \[ -\frac {30 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 4}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(30*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(sin(d*x +
c) + 1) + 15*(cos(d*x + c)^4*sin(d*x + c) + cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 10*(3*cos(d*x + c)^2 + 2
)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)

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giac [A]  time = 0.75, size = 116, normalized size = 0.97 \[ \frac {\frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, \sin \left (d x + c\right )^{2} - 38 \, \sin \left (d x + c\right ) + 25\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, \sin \left (d x + c\right )^{3} + 201 \, \sin \left (d x + c\right )^{2} + 255 \, \sin \left (d x + c\right ) + 117}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(30*log(abs(sin(d*x + c) + 1))/a - 30*log(abs(sin(d*x + c) - 1))/a + 3*(15*sin(d*x + c)^2 - 38*sin(d*x +
c) + 25)/(a*(sin(d*x + c) - 1)^2) - (55*sin(d*x + c)^3 + 201*sin(d*x + c)^2 + 255*sin(d*x + c) + 117)/(a*(sin
(d*x + c) + 1)^3))/d

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maple [A]  time = 0.17, size = 126, normalized size = 1.05 \[ \frac {1}{32 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32 a d}-\frac {1}{24 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 a d \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/a/d/(sin(d*x+c)-1)^2-1/8/a/d/(sin(d*x+c)-1)-5/32/a/d*ln(sin(d*x+c)-1)-1/24/a/d/(1+sin(d*x+c))^3-3/32/a/d/
(1+sin(d*x+c))^2-3/16/a/d/(1+sin(d*x+c))+5/32*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.30, size = 130, normalized size = 1.08 \[ -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 15 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 25 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(2*(15*sin(d*x + c)^4 + 15*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 25*sin(d*x + c) + 8)/(a*sin(d*x + c)^5 +
a*sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*a*sin(d*x + c)^2 + a*sin(d*x + c) + a) - 15*log(sin(d*x + c) + 1)/a
+ 15*log(sin(d*x + c) - 1)/a)/d

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mupad [B]  time = 0.14, size = 115, normalized size = 0.96 \[ \frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{16}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{16}-\frac {25\,{\sin \left (c+d\,x\right )}^2}{48}-\frac {25\,\sin \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atanh(sin(c + d*x)))/(16*a*d) - ((5*sin(c + d*x)^3)/16 - (25*sin(c + d*x)^2)/48 - (25*sin(c + d*x))/48 + (5
*sin(c + d*x)^4)/16 + 1/6)/(d*(a + a*sin(c + d*x) - 2*a*sin(c + d*x)^2 - 2*a*sin(c + d*x)^3 + a*sin(c + d*x)^4
+ a*sin(c + d*x)^5))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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