3.1.65 \(\int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [65]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3882, 3881, 3879} \begin {gather*} \frac {\tan (c+d x)}{5 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {\tan (c+d x)}{5 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]^2/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/5*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^3) + Tan[c + d*x]/(5*a*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(5*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]

Rule 3882

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^m/(f*(2*m + 1))), x] + Dist[m/(b*(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)]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 71, normalized size = 0.86 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (5 \sin \left (\frac {d x}{2}\right )-5 \sin \left (c+\frac {d x}{2}\right )+5 \sin \left (c+\frac {3 d x}{2}\right )+\sin \left (2 c+\frac {5 d x}{2}\right )\right )}{80 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(5*Sin[(d*x)/2] - 5*Sin[c + (d*x)/2] + 5*Sin[c + (3*d*x)/2] + Sin[2*c + (5*d*x)/2
]))/(80*a^3*d)

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Maple [A]
time = 0.08, size = 32, normalized size = 0.39

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(32\)
default	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(32\)
risch	\(\frac {2 i \left (5 \,{\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(58\)
norman	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]
time = 0.28, size = 47, normalized size = 0.57 \begin {gather*} \frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{20 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.88, size = 73, normalized size = 0.88 \begin {gather*} \frac {{\left (\cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{5 \, {\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)^2/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/5*(cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*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 ^{2}{\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)**2/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**2/(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 = 31, normalized size = 0.37 \begin {gather*} -\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{20 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.60, size = 30, normalized size = 0.36 \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 )}^4-5\right )}{20\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a + a/cos(c + d*x))^3),x)

[Out]

-(tan(c/2 + (d*x)/2)*(tan(c/2 + (d*x)/2)^4 - 5))/(20*a^3*d)

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