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

Optimal. Leaf size=82 \[ -\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac {1}{8 a d (a \sin (c+d x)+a)^2}-\frac {1}{6 d (a \sin (c+d x)+a)^3} \]

[Out]

1/8*arctanh(sin(d*x+c))/a^3/d-1/6/d/(a+a*sin(d*x+c))^3-1/8/a/d/(a+a*sin(d*x+c))^2-1/8/d/(a^3+a^3*sin(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2667, 44, 206} \[ -\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac {1}{8 a d (a \sin (c+d x)+a)^2}-\frac {1}{6 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(8*a^3*d) - 1/(6*d*(a + a*Sin[c + d*x])^3) - 1/(8*a*d*(a + a*Sin[c + d*x])^2) - 1/(8*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 (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {a \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \left (\frac {1}{2 a (a+x)^4}+\frac {1}{4 a^2 (a+x)^3}+\frac {1}{8 a^3 (a+x)^2}+\frac {1}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {1}{6 d (a+a \sin (c+d x))^3}-\frac {1}{8 a d (a+a \sin (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 a^2 d}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac {1}{6 d (a+a \sin (c+d x))^3}-\frac {1}{8 a d (a+a \sin (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 61, normalized size = 0.74 \[ \frac {-\frac {1}{8 (\sin (c+d x)+1)}-\frac {1}{8 (\sin (c+d x)+1)^2}-\frac {1}{6 (\sin (c+d x)+1)^3}+\frac {1}{8} \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[c + d*x]]/8 - 1/(6*(1 + Sin[c + d*x])^3) - 1/(8*(1 + Sin[c + d*x])^2) - 1/(8*(1 + Sin[c + d*x])))
/(a^3*d)

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fricas [B]  time = 0.82, size = 154, normalized size = 1.88 \[ -\frac {6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, \sin \left (d x + c\right ) - 26}{48 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(6*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) +
3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(-sin(d*x + c) + 1) - 18*sin(d*x + c) - 26)/(
3*a^3*d*cos(d*x + c)^2 - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 4*a^3*d)*sin(d*x + c))

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giac [A]  time = 1.44, size = 81, normalized size = 0.99 \[ \frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} - \frac {11 \, \sin \left (d x + c\right )^{3} + 45 \, \sin \left (d x + c\right )^{2} + 69 \, \sin \left (d x + c\right ) + 51}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(6*log(abs(sin(d*x + c) + 1))/a^3 - 6*log(abs(sin(d*x + c) - 1))/a^3 - (11*sin(d*x + c)^3 + 45*sin(d*x +
c)^2 + 69*sin(d*x + c) + 51)/(a^3*(sin(d*x + c) + 1)^3))/d

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maple [A]  time = 0.21, size = 90, normalized size = 1.10 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{16 a^{3} d}-\frac {1}{6 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 a^{3} d \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{16 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a^3/d*ln(sin(d*x+c)-1)-1/6/a^3/d/(1+sin(d*x+c))^3-1/8/a^3/d/(1+sin(d*x+c))^2-1/8/a^3/d/(1+sin(d*x+c))+1/
16*ln(1+sin(d*x+c))/a^3/d

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maxima [A]  time = 0.50, size = 98, normalized size = 1.20 \[ -\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 10\right )}}{a^{3} \sin \left (d x + c\right )^{3} + 3 \, a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(2*(3*sin(d*x + c)^2 + 9*sin(d*x + c) + 10)/(a^3*sin(d*x + c)^3 + 3*a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x +
c) + a^3) - 3*log(sin(d*x + c) + 1)/a^3 + 3*log(sin(d*x + c) - 1)/a^3)/d

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mupad [B]  time = 4.59, size = 83, normalized size = 1.01 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{8\,a^3\,d}-\frac {\frac {{\sin \left (c+d\,x\right )}^2}{8}+\frac {3\,\sin \left (c+d\,x\right )}{8}+\frac {5}{12}}{d\,\left (a^3\,{\sin \left (c+d\,x\right )}^3+3\,a^3\,{\sin \left (c+d\,x\right )}^2+3\,a^3\,\sin \left (c+d\,x\right )+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(sin(c + d*x))/(8*a^3*d) - ((3*sin(c + d*x))/8 + sin(c + d*x)^2/8 + 5/12)/(d*(3*a^3*sin(c + d*x) + a^3 +
3*a^3*sin(c + d*x)^2 + a^3*sin(c + d*x)^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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