3.1.67 \(\int \frac {1}{(a+a \sec (c+d x))^3} \, dx\) [67]

Optimal. Leaf size=88 \[ \frac {x}{a^3}-\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {22 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

x/a^3-1/5*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-7/15*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-22/15*tan(d*x+c)/d/(a^3+a^3*s
ec(d*x+c))

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Rubi [A]
time = 0.08, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3862, 4007, 4004, 3879} \begin {gather*} -\frac {22 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x}{a^3}-\frac {7 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(-3),x]

[Out]

x/a^3 - Tan[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) - (7*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (22*Tan
[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 3862

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*
(2*n + 1))), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*
x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4007

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-(b
*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[
e + f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f
}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-5 a+2 a \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {15 a^2-7 a^2 \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac {x}{a^3}-\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {22 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac {x}{a^3}-\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {22 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 162, normalized size = 1.84 \begin {gather*} \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (60 d x \cos ^5\left (\frac {1}{2} (c+d x)\right )-3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+26 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-128 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-3 \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+26 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{15 a^3 d (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(60*d*x*Cos[(c + d*x)/2]^5 - 3*Sec[c/2]*Sin[(d*x)/2] + 26*Cos[(c + d*x)/2]^
2*Sec[c/2]*Sin[(d*x)/2] - 128*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] - 3*Cos[(c + d*x)/2]*Tan[c/2] + 26*Cos[
(c + d*x)/2]^3*Tan[c/2]))/(15*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.06, size = 59, normalized size = 0.67

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(59\)
default	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(59\)
norman	\(\frac {\frac {x}{a}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a d}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}}{a^{2}}\)	\(66\)
risch	\(\frac {x}{a^{3}}-\frac {2 i \left (45 \,{\mathrm e}^{4 i \left (d x +c \right )}+135 \,{\mathrm e}^{3 i \left (d x +c \right )}+185 \,{\mathrm e}^{2 i \left (d x +c \right )}+115 \,{\mathrm e}^{i \left (d x +c \right )}+32\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(-1/5*tan(1/2*d*x+1/2*c)^5+4/3*tan(1/2*d*x+1/2*c)^3-7*tan(1/2*d*x+1/2*c)+8*arctan(tan(1/2*d*x+1/2*c)
))

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Maxima [A]
time = 0.50, size = 92, normalized size = 1.05 \begin {gather*} -\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]
time = 1.89, size = 116, normalized size = 1.32 \begin {gather*} \frac {15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (32 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 22\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (32*cos(d*x + c)^2 + 51*c
os(d*x + c) + 22)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))**3,x)

[Out]

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

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Giac [A]
time = 0.43, size = 68, normalized size = 0.77 \begin {gather*} \frac {\frac {60 \, {\left (d x + c\right )}}{a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)/a^3 - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d

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Mupad [B]
time = 0.69, size = 81, normalized size = 0.92 \begin {gather*} \frac {x}{a^3}-\frac {\frac {32\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-\frac {13\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{30}+\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a/cos(c + d*x))^3,x)

[Out]

x/a^3 - (sin(c/2 + (d*x)/2)/20 - (13*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2))/30 + (32*cos(c/2 + (d*x)/2)^4*si
n(c/2 + (d*x)/2))/15)/(a^3*d*cos(c/2 + (d*x)/2)^5)

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