3.1.75 \(\int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx\) [75]

Optimal. Leaf size=112 \[ -\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {8 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {8 \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \]

[Out]

-1/7*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+4/35*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3+8/105*tan(d*x+c)/d/(a^2+a^2*sec(d*
x+c))^2+8/105*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))

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Rubi [A]
time = 0.09, antiderivative size = 112, 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 = {3882, 3881, 3879} \begin {gather*} \frac {8 \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac {8 \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {4 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^4,x]

[Out]

-1/7*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^4) + (4*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) + (8*Tan[c + d
*x])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + (8*Tan[c + d*x])/(105*d*(a^4 + a^4*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))^4} \, dx &=-\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {4 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx}{7 a}\\ &=-\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {8 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^2}\\ &=-\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {8 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {8 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=-\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {8 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {8 \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 99, normalized size = 0.88 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (280 \sin \left (\frac {d x}{2}\right )-175 \sin \left (c+\frac {d x}{2}\right )+168 \sin \left (c+\frac {3 d x}{2}\right )-105 \sin \left (2 c+\frac {3 d x}{2}\right )+91 \sin \left (2 c+\frac {5 d x}{2}\right )+13 \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(280*Sin[(d*x)/2] - 175*Sin[c + (d*x)/2] + 168*Sin[c + (3*d*x)/2] - 105*Sin[2*c +
(3*d*x)/2] + 91*Sin[2*c + (5*d*x)/2] + 13*Sin[3*c + (7*d*x)/2]))/(6720*a^4*d)

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Maple [A]
time = 0.08, size = 58, normalized size = 0.52

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
default	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
risch	\(\frac {2 i \left (105 \,{\mathrm e}^{5 i \left (d x +c \right )}+175 \,{\mathrm e}^{4 i \left (d x +c \right )}+280 \,{\mathrm e}^{3 i \left (d x +c \right )}+168 \,{\mathrm e}^{2 i \left (d x +c \right )}+91 \,{\mathrm e}^{i \left (d x +c \right )}+13\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(80\)
norman	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a d}-\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{70 a d}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/8/d/a^4*(1/7*tan(1/2*d*x+1/2*c)^7-1/5*tan(1/2*d*x+1/2*c)^5-1/3*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 = 87, normalized size = 0.78 \begin {gather*} \frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} 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))^4,x, algorithm="maxima")

[Out]

1/840*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4*d)

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Fricas [A]
time = 2.20, size = 99, normalized size = 0.88 \begin {gather*} \frac {{\left (13 \, \cos \left (d x + c\right )^{3} + 52 \, \cos \left (d x + c\right )^{2} + 32 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/105*(13*cos(d*x + c)^3 + 52*cos(d*x + c)^2 + 32*cos(d*x + c) + 8)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4
*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*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 ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \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))**4,x)

[Out]

Integral(sec(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)/a*
*4

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Giac [A]
time = 0.45, size = 59, normalized size = 0.53 \begin {gather*} \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} 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))^4,x, algorithm="giac")

[Out]

1/840*(15*tan(1/2*d*x + 1/2*c)^7 - 21*tan(1/2*d*x + 1/2*c)^5 - 35*tan(1/2*d*x + 1/2*c)^3 + 105*tan(1/2*d*x + 1
/2*c))/(a^4*d)

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Mupad [B]
time = 0.67, size = 58, normalized size = 0.52 \begin {gather*} -\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-105\right )}{840\,a^4\,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))^4),x)

[Out]

-(tan(c/2 + (d*x)/2)*(35*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^4 - 15*tan(c/2 + (d*x)/2)^6 - 105))/(840
*a^4*d)

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