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

Optimal. Leaf size=126 \[ \frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a \sin (c+d x)+a)^4}-\frac {1}{12 d (a \sin (c+d x)+a)^3}-\frac {3}{32 a d (a \sin (c+d x)+a)^2} \]

[Out]

5/32*arctanh(sin(d*x+c))/a^3/d-1/16*a/d/(a+a*sin(d*x+c))^4-1/12/d/(a+a*sin(d*x+c))^3-3/32/a/d/(a+a*sin(d*x+c))
^2+1/32/d/(a^3-a^3*sin(d*x+c))-1/8/d/(a^3+a^3*sin(d*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 126, 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} \[ \frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a \sin (c+d x)+a)^4}-\frac {1}{12 d (a \sin (c+d x)+a)^3}-\frac {3}{32 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sin[c + d*x]])/(32*a^3*d) - a/(16*d*(a + a*Sin[c + d*x])^4) - 1/(12*d*(a + a*Sin[c + d*x])^3) - 3/(
32*a*d*(a + a*Sin[c + d*x])^2) + 1/(32*d*(a^3 - a^3*Sin[c + d*x])) - 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 ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^2}+\frac {1}{4 a^2 (a+x)^5}+\frac {1}{4 a^3 (a+x)^4}+\frac {3}{16 a^4 (a+x)^3}+\frac {1}{8 a^5 (a+x)^2}+\frac {5}{32 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{12 d (a+a \sin (c+d x))^3}-\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{32 a^2 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{12 d (a+a \sin (c+d x))^3}-\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 95, normalized size = 0.75 \[ -\frac {\sec ^2(c+d x) \left (-15 \sin ^4(c+d x)-45 \sin ^3(c+d x)-35 \sin ^2(c+d x)+15 \sin (c+d x)+15 (\sin (c+d x)-1) (\sin (c+d x)+1)^4 \tanh ^{-1}(\sin (c+d x))+32\right )}{96 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(Sec[c + d*x]^2*(32 + 15*Sin[c + d*x] - 35*Sin[c + d*x]^2 - 45*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 15*A
rcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])*(1 + Sin[c + d*x])^4))/(a^3*d*(1 + Sin[c + d*x])^3)

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fricas [A]  time = 0.61, size = 228, normalized size = 1.81 \[ -\frac {30 \, \cos \left (d x + c\right )^{4} - 130 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 30 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 36}{192 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(30*cos(d*x + c)^4 - 130*cos(d*x + c)^2 - 15*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 4
*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) + 15*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)
^4 - 4*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 30*(3*cos(d*x + c)^2 - 2)*sin(d*x + c) + 36)/(3*
a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + (a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2)*sin(d*x + c))

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giac [A]  time = 0.74, size = 116, normalized size = 0.92 \[ \frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac {12 \, {\left (5 \, \sin \left (d x + c\right ) - 7\right )}}{a^{3} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 596 \, \sin \left (d x + c\right )^{3} + 1110 \, \sin \left (d x + c\right )^{2} + 996 \, \sin \left (d x + c\right ) + 405}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(60*log(abs(sin(d*x + c) + 1))/a^3 - 60*log(abs(sin(d*x + c) - 1))/a^3 + 12*(5*sin(d*x + c) - 7)/(a^3*(s
in(d*x + c) - 1)) - (125*sin(d*x + c)^4 + 596*sin(d*x + c)^3 + 1110*sin(d*x + c)^2 + 996*sin(d*x + c) + 405)/(
a^3*(sin(d*x + c) + 1)^4))/d

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maple [A]  time = 0.26, size = 126, normalized size = 1.00 \[ -\frac {1}{32 a^{3} d \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{64 a^{3} d}-\frac {1}{16 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 a^{3} d \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{64 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/a^3/d/(sin(d*x+c)-1)-5/64/a^3/d*ln(sin(d*x+c)-1)-1/16/a^3/d/(1+sin(d*x+c))^4-1/12/a^3/d/(1+sin(d*x+c))^3
-3/32/a^3/d/(1+sin(d*x+c))^2-1/8/a^3/d/(1+sin(d*x+c))+5/64*ln(1+sin(d*x+c))/a^3/d

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maxima [A]  time = 0.33, size = 146, normalized size = 1.16 \[ -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 45 \, \sin \left (d x + c\right )^{3} + 35 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) - 32\right )}}{a^{3} \sin \left (d x + c\right )^{5} + 3 \, a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - a^{3}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(2*(15*sin(d*x + c)^4 + 45*sin(d*x + c)^3 + 35*sin(d*x + c)^2 - 15*sin(d*x + c) - 32)/(a^3*sin(d*x + c)
^5 + 3*a^3*sin(d*x + c)^4 + 2*a^3*sin(d*x + c)^3 - 2*a^3*sin(d*x + c)^2 - 3*a^3*sin(d*x + c) - a^3) - 15*log(s
in(d*x + c) + 1)/a^3 + 15*log(sin(d*x + c) - 1)/a^3)/d

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mupad [B]  time = 0.15, size = 129, normalized size = 1.02 \[ \frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{32\,a^3\,d}+\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {15\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {35\,{\sin \left (c+d\,x\right )}^2}{96}-\frac {5\,\sin \left (c+d\,x\right )}{32}-\frac {1}{3}}{d\,\left (-a^3\,{\sin \left (c+d\,x\right )}^5-3\,a^3\,{\sin \left (c+d\,x\right )}^4-2\,a^3\,{\sin \left (c+d\,x\right )}^3+2\,a^3\,{\sin \left (c+d\,x\right )}^2+3\,a^3\,\sin \left (c+d\,x\right )+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atanh(sin(c + d*x)))/(32*a^3*d) + ((35*sin(c + d*x)^2)/96 - (5*sin(c + d*x))/32 + (15*sin(c + d*x)^3)/32 +
(5*sin(c + d*x)^4)/32 - 1/3)/(d*(3*a^3*sin(c + d*x) + a^3 + 2*a^3*sin(c + d*x)^2 - 2*a^3*sin(c + d*x)^3 - 3*a^
3*sin(c + d*x)^4 - a^3*sin(c + d*x)^5))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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