3.1.85 \(\int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx\) [85]

Optimal. Leaf size=143 \[ -\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{63 a d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {2 \tan (c+d x)}{63 d \left (a^5+a^5 \sec (c+d x)\right )} \]

[Out]

-1/9*tan(d*x+c)/d/(a+a*sec(d*x+c))^5+5/63*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^4+1/21*tan(d*x+c)/a^2/d/(a+a*sec(d*x
+c))^3+2/63*tan(d*x+c)/a/d/(a^2+a^2*sec(d*x+c))^2+2/63*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))

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Rubi [A]
time = 0.11, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {2 \tan (c+d x)}{63 d \left (a^5 \sec (c+d x)+a^5\right )}+\frac {2 \tan (c+d x)}{63 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {\tan (c+d x)}{21 a^2 d (a \sec (c+d x)+a)^3}+\frac {5 \tan (c+d x)}{63 a d (a \sec (c+d x)+a)^4}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^5,x]

[Out]

-1/9*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^5) + (5*Tan[c + d*x])/(63*a*d*(a + a*Sec[c + d*x])^4) + Tan[c + d*x]
/(21*a^2*d*(a + a*Sec[c + d*x])^3) + (2*Tan[c + d*x])/(63*a*d*(a^2 + a^2*Sec[c + d*x])^2) + (2*Tan[c + d*x])/(
63*d*(a^5 + a^5*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))^5} \, dx &=-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx}{9 a}\\ &=-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {5 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx}{21 a^2}\\ &=-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac {2 \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{21 a^3}\\ &=-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {2 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{63 a^4}\\ &=-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {2 \tan (c+d x)}{63 d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 125, normalized size = 0.87 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (315 \sin \left (\frac {d x}{2}\right )-315 \sin \left (c+\frac {d x}{2}\right )+273 \sin \left (c+\frac {3 d x}{2}\right )-147 \sin \left (2 c+\frac {3 d x}{2}\right )+117 \sin \left (2 c+\frac {5 d x}{2}\right )-63 \sin \left (3 c+\frac {5 d x}{2}\right )+45 \sin \left (3 c+\frac {7 d x}{2}\right )+5 \sin \left (4 c+\frac {9 d x}{2}\right )\right )}{16128 a^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^5,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(315*Sin[(d*x)/2] - 315*Sin[c + (d*x)/2] + 273*Sin[c + (3*d*x)/2] - 147*Sin[2*c +
(3*d*x)/2] + 117*Sin[2*c + (5*d*x)/2] - 63*Sin[3*c + (5*d*x)/2] + 45*Sin[3*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*
x)/2]))/(16128*a^5*d)

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Maple [A]
time = 0.08, size = 58, normalized size = 0.41

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}+\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(58\)
default	\(\frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}+\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(58\)
risch	\(\frac {2 i \left (63 \,{\mathrm e}^{7 i \left (d x +c \right )}+147 \,{\mathrm e}^{6 i \left (d x +c \right )}+315 \,{\mathrm e}^{5 i \left (d x +c \right )}+315 \,{\mathrm e}^{4 i \left (d x +c \right )}+273 \,{\mathrm e}^{3 i \left (d x +c \right )}+117 \,{\mathrm e}^{2 i \left (d x +c \right )}+45 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\)	\(102\)
norman	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a d}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a d}+\frac {25 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1008 a d}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 a d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+a*sec(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

1/16/d/a^5*(-1/9*tan(1/2*d*x+1/2*c)^9+2/7*tan(1/2*d*x+1/2*c)^7-2/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]
time = 0.30, size = 87, normalized size = 0.61 \begin {gather*} \frac {\frac {63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} 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))^5,x, algorithm="maxima")

[Out]

1/1008*(63*sin(d*x + c)/(cos(d*x + c) + 1) - 42*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 18*sin(d*x + c)^7/(cos(d
*x + c) + 1)^7 - 7*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)

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Fricas [A]
time = 15.42, size = 123, normalized size = 0.86 \begin {gather*} \frac {{\left (5 \, \cos \left (d x + c\right )^{4} + 25 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} 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))^5,x, algorithm="fricas")

[Out]

1/63*(5*cos(d*x + c)^4 + 25*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 10*cos(d*x + c) + 2)*sin(d*x + c)/(a^5*d*cos(
d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c)
+ a^5*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 ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \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))**5,x)

[Out]

Integral(sec(c + d*x)**2/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d*x)**3 + 10*sec(c + d*x)**2 + 5*se
c(c + d*x) + 1), x)/a**5

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Giac [A]
time = 0.50, size = 59, normalized size = 0.41 \begin {gather*} -\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1008 \, a^{5} 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))^5,x, algorithm="giac")

[Out]

-1/1008*(7*tan(1/2*d*x + 1/2*c)^9 - 18*tan(1/2*d*x + 1/2*c)^7 + 42*tan(1/2*d*x + 1/2*c)^3 - 63*tan(1/2*d*x + 1
/2*c))/(a^5*d)

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Mupad [B]
time = 0.69, size = 58, normalized size = 0.41 \begin {gather*} -\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-63\right )}{1008\,a^5\,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))^5),x)

[Out]

-(tan(c/2 + (d*x)/2)*(42*tan(c/2 + (d*x)/2)^2 - 18*tan(c/2 + (d*x)/2)^6 + 7*tan(c/2 + (d*x)/2)^8 - 63))/(1008*
a^5*d)

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