3.1.56 \(\int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [56]

Optimal. Leaf size=55 \[ \frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, 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 {\tan (c+d x)}{3 d \left (a^2 \sec (c+d x)+a^2\right )}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + a*Sec[c + d*x])^2,x]

[Out]

Tan[c + d*x]/(3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*d*(a^2 + a^2*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))^2} \, dx &=\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 60, normalized size = 1.09 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (3 \sin \left (\frac {d x}{2}\right )-3 \sin \left (c+\frac {d x}{2}\right )+2 \sin \left (c+\frac {3 d x}{2}\right )\right )}{12 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a + a*Sec[c + d*x])^2,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(3*Sin[(d*x)/2] - 3*Sin[c + (d*x)/2] + 2*Sin[c + (3*d*x)/2]))/(12*a^2*d)

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Maple [A]
time = 0.05, size = 32, normalized size = 0.58

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(32\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(32\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}}{a}\)	\(42\)
risch	\(\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}+2\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2/d/a^2*(-1/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]
time = 0.28, size = 47, normalized size = 0.85 \begin {gather*} \frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{6 \, a^{2} 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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.41, size = 51, normalized size = 0.93 \begin {gather*} \frac {{\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/3*(2*cos(d*x + c) + 1)*sin(d*x + c)/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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 ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+a*sec(d*x+c))**2,x)

[Out]

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

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Giac [A]
time = 0.46, size = 31, normalized size = 0.56 \begin {gather*} -\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{6 \, a^{2} 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))^2,x, algorithm="giac")

[Out]

-1/6*(tan(1/2*d*x + 1/2*c)^3 - 3*tan(1/2*d*x + 1/2*c))/(a^2*d)

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Mupad [B]
time = 0.60, size = 30, normalized size = 0.55 \begin {gather*} -\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3\right )}{6\,a^2\,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))^2),x)

[Out]

-(tan(c/2 + (d*x)/2)*(tan(c/2 + (d*x)/2)^2 - 3))/(6*a^2*d)

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