3.1.66 \(\int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [66]

Optimal. Leaf size=83 \[ \frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {2 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

1/5*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+2/15*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+2/15*tan(d*x+c)/d/(a^3+a^3*sec(d*x+
c))

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Rubi [A]
time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3881, 3879} \begin {gather*} \frac {2 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {2 \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[Sec[c + d*x]/(a + a*Sec[c + d*x])^3,x]

[Out]

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

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 3881

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a}\\ &=\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {2 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {2 \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 = 0.24, size = 86, normalized size = 1.04 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (40 \sin \left (\frac {d x}{2}\right )-30 \sin \left (c+\frac {d x}{2}\right )+20 \sin \left (c+\frac {3 d x}{2}\right )-15 \sin \left (2 c+\frac {3 d x}{2}\right )+7 \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{240 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(40*Sin[(d*x)/2] - 30*Sin[c + (d*x)/2] + 20*Sin[c + (3*d*x)/2] - 15*Sin[2*c + (3*
d*x)/2] + 7*Sin[2*c + (5*d*x)/2]))/(240*a^3*d)

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Maple [A]
time = 0.06, size = 45, normalized size = 0.54

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(45\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(45\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}}{a^{2}}\)	\(61\)
risch	\(\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{3 i \left (d x +c \right )}+40 \,{\mathrm e}^{2 i \left (d x +c \right )}+20 \,{\mathrm e}^{i \left (d x +c \right )}+7\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(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)^5-2/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]
time = 0.29, size = 67, normalized size = 0.81 \begin {gather*} \frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \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}}}{60 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*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*d)

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Fricas [A]
time = 2.50, size = 75, normalized size = 0.90 \begin {gather*} \frac {{\left (7 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2\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(sec(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(7*cos(d*x + c)^2 + 6*cos(d*x + c) + 2)*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 {\sec {\left (c + d x \right )}}{\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(sec(d*x+c)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)/(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 = 0.48, size = 46, normalized size = 0.55 \begin {gather*} \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*tan(1/2*d*x + 1/2*c)^5 - 10*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d*x + 1/2*c))/(a^3*d)

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Mupad [B]
time = 0.62, size = 45, normalized size = 0.54 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\right )}{60\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 + 15))/(60*a^3*d)

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