3.468 \(\int \frac {\sec ^8(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=127 \[ \frac {(5 a+b) (a-b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac {(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}-\frac {(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 205} \[ \frac {(5 a+b) (a-b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac {(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}-\frac {(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((a - b)^2*(5*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*b^(7/2)*d) - ((2*a - 3*b)*Tan[c + d*x]
)/(b^3*d) + Tan[c + d*x]^3/(3*b^2*d) - ((a - b)^3*Tan[c + d*x])/(2*a*b^3*d*(a + b*Tan[c + d*x]^2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {2 a-3 b}{b^3}+\frac {x^2}{b^2}+\frac {(a-b)^2 (2 a+b)+3 (a-b)^2 b x^2}{b^3 \left (a+b x^2\right )^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d}+\frac {\operatorname {Subst}\left (\int \frac {(a-b)^2 (2 a+b)+3 (a-b)^2 b x^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}\\ &=-\frac {(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d}-\frac {(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}+\frac {\left ((a-b)^2 (5 a+b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a b^3 d}\\ &=\frac {(a-b)^2 (5 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac {(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d}-\frac {(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 135, normalized size = 1.06 \[ \frac {\frac {3 (a-b)^2 (5 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+4 \sqrt {b} (4 b-3 a) \tan (c+d x)+\frac {3 \sqrt {b} (b-a)^3 \sin (2 (c+d x))}{a ((a-b) \cos (2 (c+d x))+a+b)}+2 b^{3/2} \tan (c+d x) \sec ^2(c+d x)}{6 b^{7/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((3*(a - b)^2*(5*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2) + (3*Sqrt[b]*(-a + b)^3*Sin[2*(c + d*x
)])/(a*(a + b + (a - b)*Cos[2*(c + d*x)])) + 4*Sqrt[b]*(-3*a + 4*b)*Tan[c + d*x] + 2*b^(3/2)*Sec[c + d*x]^2*Ta
n[c + d*x])/(6*b^(7/2)*d)

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fricas [B]  time = 0.62, size = 597, normalized size = 4.70 \[ \left [-\frac {3 \, {\left ({\left (5 \, a^{4} - 14 \, a^{3} b + 12 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (5 \, a^{3} b - 9 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a b} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 4 \, {\left (2 \, a^{2} b^{3} - {\left (15 \, a^{4} b - 37 \, a^{3} b^{2} + 25 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} d \cos \left (d x + c\right )^{3} + {\left (a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{5}\right )}}, -\frac {3 \, {\left ({\left (5 \, a^{4} - 14 \, a^{3} b + 12 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (5 \, a^{3} b - 9 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) - 2 \, {\left (2 \, a^{2} b^{3} - {\left (15 \, a^{4} b - 37 \, a^{3} b^{2} + 25 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{5} d \cos \left (d x + c\right )^{3} + {\left (a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[-1/24*(3*((5*a^4 - 14*a^3*b + 12*a^2*b^2 - 2*a*b^3 - b^4)*cos(d*x + c)^5 + (5*a^3*b - 9*a^2*b^2 + 3*a*b^3 + b
^4)*cos(d*x + c)^3)*sqrt(-a*b)*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 + 4*((
a + b)*cos(d*x + c)^3 - b*cos(d*x + c))*sqrt(-a*b)*sin(d*x + c) + b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2
*(a*b - b^2)*cos(d*x + c)^2 + b^2)) - 4*(2*a^2*b^3 - (15*a^4*b - 37*a^3*b^2 + 25*a^2*b^3 - 3*a*b^4)*cos(d*x +
c)^4 - 2*(5*a^3*b^2 - 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^5*d*cos(d*x + c)^3 + (a^3*b^4 - a^2*b^5)
*d*cos(d*x + c)^5), -1/12*(3*((5*a^4 - 14*a^3*b + 12*a^2*b^2 - 2*a*b^3 - b^4)*cos(d*x + c)^5 + (5*a^3*b - 9*a^
2*b^2 + 3*a*b^3 + b^4)*cos(d*x + c)^3)*sqrt(a*b)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)*sqrt(a*b)/(a*b*cos(d*
x + c)*sin(d*x + c))) - 2*(2*a^2*b^3 - (15*a^4*b - 37*a^3*b^2 + 25*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^4 - 2*(5*a^
3*b^2 - 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^5*d*cos(d*x + c)^3 + (a^3*b^4 - a^2*b^5)*d*cos(d*x + c
)^5)]

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giac [A]  time = 2.29, size = 180, normalized size = 1.42 \[ \frac {\frac {3 \, {\left (5 \, a^{3} - 9 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} a b^{3}} - \frac {3 \, {\left (a^{3} \tan \left (d x + c\right ) - 3 \, a^{2} b \tan \left (d x + c\right ) + 3 \, a b^{2} \tan \left (d x + c\right ) - b^{3} \tan \left (d x + c\right )\right )}}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b^{3}} + \frac {2 \, {\left (b^{4} \tan \left (d x + c\right )^{3} - 6 \, a b^{3} \tan \left (d x + c\right ) + 9 \, b^{4} \tan \left (d x + c\right )\right )}}{b^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/6*(3*(5*a^3 - 9*a^2*b + 3*a*b^2 + b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b
)))/(sqrt(a*b)*a*b^3) - 3*(a^3*tan(d*x + c) - 3*a^2*b*tan(d*x + c) + 3*a*b^2*tan(d*x + c) - b^3*tan(d*x + c))/
((b*tan(d*x + c)^2 + a)*a*b^3) + 2*(b^4*tan(d*x + c)^3 - 6*a*b^3*tan(d*x + c) + 9*b^4*tan(d*x + c))/b^6)/d

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maple [B]  time = 0.92, size = 275, normalized size = 2.17 \[ \frac {\tan ^{3}\left (d x +c \right )}{3 b^{2} d}-\frac {2 a \tan \left (d x +c \right )}{d \,b^{3}}+\frac {3 \tan \left (d x +c \right )}{b^{2} d}-\frac {a^{2} \tan \left (d x +c \right )}{2 d \,b^{3} \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {3 a \tan \left (d x +c \right )}{2 d \,b^{2} \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}-\frac {3 \tan \left (d x +c \right )}{2 d b \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {\tan \left (d x +c \right )}{2 a d \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {5 a^{2} \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \,b^{3} \sqrt {a b}}-\frac {9 a \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \,b^{2} \sqrt {a b}}+\frac {3 \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d b \sqrt {a b}}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d a \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8/(a+b*tan(d*x+c)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.81, size = 137, normalized size = 1.08 \[ -\frac {\frac {3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (d x + c\right )}{a b^{4} \tan \left (d x + c\right )^{2} + a^{2} b^{3}} - \frac {2 \, {\left (b \tan \left (d x + c\right )^{3} - 3 \, {\left (2 \, a - 3 \, b\right )} \tan \left (d x + c\right )\right )}}{b^{3}} - \frac {3 \, {\left (5 \, a^{3} - 9 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/6*(3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*tan(d*x + c)/(a*b^4*tan(d*x + c)^2 + a^2*b^3) - 2*(b*tan(d*x + c)^3 -
3*(2*a - 3*b)*tan(d*x + c))/b^3 - 3*(5*a^3 - 9*a^2*b + 3*a*b^2 + b^3)*arctan(b*tan(d*x + c)/sqrt(a*b))/(sqrt(a
*b)*a*b^3))/d

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mupad [B]  time = 12.17, size = 167, normalized size = 1.31 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {2\,a}{b^3}-\frac {3}{b^2}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,a\,d\,\left (b^4\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b^3\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^2\,\left (5\,a+b\right )}{\sqrt {a}\,\left (5\,a^3-9\,a^2\,b+3\,a\,b^2+b^3\right )}\right )\,{\left (a-b\right )}^2\,\left (5\,a+b\right )}{2\,a^{3/2}\,b^{7/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^8*(a + b*tan(c + d*x)^2)^2),x)

[Out]

tan(c + d*x)^3/(3*b^2*d) - (tan(c + d*x)*((2*a)/b^3 - 3/b^2))/d - (tan(c + d*x)*(3*a*b^2 - 3*a^2*b + a^3 - b^3
))/(2*a*d*(a*b^3 + b^4*tan(c + d*x)^2)) + (atan((b^(1/2)*tan(c + d*x)*(a - b)^2*(5*a + b))/(a^(1/2)*(3*a*b^2 -
9*a^2*b + 5*a^3 + b^3)))*(a - b)^2*(5*a + b))/(2*a^(3/2)*b^(7/2)*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

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