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

Optimal. Leaf size=108 \[ \frac {\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac {(a-b)^3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2} d}-\frac {(a-3 b) \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^5(c+d x)}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3675, 390, 205} \[ \frac {\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac {(a-3 b) \tan ^3(c+d x)}{3 b^2 d}-\frac {(a-b)^3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2} d}+\frac {\tan ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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)}{a+b \tan ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2-3 a b+3 b^2}{b^3}-\frac {(a-3 b) x^2}{b^2}+\frac {x^4}{b}+\frac {-a^3+3 a^2 b-3 a b^2+b^3}{b^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac {(a-3 b) \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^5(c+d x)}{5 b d}-\frac {(a-b)^3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{b^3 d}\\ &=-\frac {(a-b)^3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2} d}+\frac {\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac {(a-3 b) \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^5(c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 103, normalized size = 0.95 \[ \frac {\sqrt {b} \tan (c+d x) \left (15 a^2-b (5 a-9 b) \sec ^2(c+d x)-40 a b+3 b^2 \sec ^4(c+d x)+33 b^2\right )-\frac {15 (a-b)^3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a}}}{15 b^{7/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-15*(a - b)^3*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a] + Sqrt[b]*(15*a^2 - 40*a*b + 33*b^2 - (5*a - 9
*b)*b*Sec[c + d*x]^2 + 3*b^2*Sec[c + d*x]^4)*Tan[c + d*x])/(15*b^(7/2)*d)

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fricas [A]  time = 0.56, size = 425, normalized size = 3.94 \[ \left [\frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {-a b} \cos \left (d x + c\right )^{5} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (15 \, a^{3} b - 40 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b^{3} - {\left (5 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, a b^{4} d \cos \left (d x + c\right )^{5}}, \frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left ({\left (15 \, a^{3} b - 40 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b^{3} - {\left (5 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, a b^{4} d \cos \left (d x + c\right )^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(-a*b)*cos(d*x + c)^5*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*((15*a^3*b - 40*a^2*b^2 + 33*a*b^
3)*cos(d*x + c)^4 + 3*a*b^3 - (5*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^5), 1/
30*(15*(a^3 - 3*a^2*b + 3*a*b^2 - b^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)))*cos(d*x + c)^5 + 2*((15*a^3*b - 40*a^2*b^2 + 33*a*b^3)*cos(d*x + c)^4 + 3*a*b^3 - (5*a^2
*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^5)]

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giac [A]  time = 1.43, size = 151, normalized size = 1.40 \[ -\frac {\frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} b^{3}} - \frac {3 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a b^{3} \tan \left (d x + c\right )^{3} + 15 \, b^{4} \tan \left (d x + c\right )^{3} + 15 \, a^{2} b^{2} \tan \left (d x + c\right ) - 45 \, a b^{3} \tan \left (d x + c\right ) + 45 \, b^{4} \tan \left (d x + c\right )}{b^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*(a^3 - 3*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)*b^3) - (3*b^4*tan(d*x + c)^5 - 5*a*b^3*tan(d*x + c)^3 + 15*b^4*tan(d*x + c)^3 + 15*a^2*b^2*tan
(d*x + c) - 45*a*b^3*tan(d*x + c) + 45*b^4*tan(d*x + c))/b^5)/d

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maple [B]  time = 0.68, size = 206, normalized size = 1.91 \[ \frac {\tan ^{5}\left (d x +c \right )}{5 b d}-\frac {a \left (\tan ^{3}\left (d x +c \right )\right )}{3 b^{2} d}+\frac {\tan ^{3}\left (d x +c \right )}{b d}+\frac {a^{2} \tan \left (d x +c \right )}{d \,b^{3}}-\frac {3 a \tan \left (d x +c \right )}{b^{2} d}+\frac {3 \tan \left (d x +c \right )}{b d}-\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right ) a^{3}}{d \,b^{3} \sqrt {a b}}+\frac {3 \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right ) a^{2}}{d \,b^{2} \sqrt {a b}}-\frac {3 \arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right ) a}{d b \sqrt {a b}}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{d \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 110, normalized size = 1.02 \[ -\frac {\frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{2} \tan \left (d x + c\right )^{5} - 5 \, {\left (a b - 3 \, b^{2}\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (a^{2} - 3 \, a b + 3 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{3}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*arctan(b*tan(d*x + c)/sqrt(a*b))/(sqrt(a*b)*b^3) - (3*b^2*tan(d*x +
c)^5 - 5*(a*b - 3*b^2)*tan(d*x + c)^3 + 15*(a^2 - 3*a*b + 3*b^2)*tan(d*x + c))/b^3)/d

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mupad [B]  time = 12.29, size = 136, normalized size = 1.26 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {3}{b}+\frac {a\,\left (\frac {a}{b^2}-\frac {3}{b}\right )}{b}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,b\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a}{3\,b^2}-\frac {1}{b}\right )}{d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^3}{\sqrt {a}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}\right )\,{\left (a-b\right )}^3}{\sqrt {a}\,b^{7/2}\,d} \]

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)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{8}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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