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

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

[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)

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Rubi [A]  time = 0.109796, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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])

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 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]

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*}

Mathematica [A]  time = 0.922971, 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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Maple [B]  time = 0.073, size = 206, normalized size = 1.9 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5\,bd}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}a}{3\,d{b}^{2}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{bd}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d{b}^{3}}}-3\,{\frac{a\tan \left ( dx+c \right ) }{d{b}^{2}}}+3\,{\frac{\tan \left ( dx+c \right ) }{bd}}-{\frac{{a}^{3}}{d{b}^{3}}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+3\,{\frac{{a}^{2}}{d{b}^{2}\sqrt{ab}}\arctan \left ({\frac{b\tan \left ( dx+c \right ) }{\sqrt{ab}}} \right ) }-3\,{\frac{a}{bd\sqrt{ab}}\arctan \left ({\frac{b\tan \left ( dx+c \right ) }{\sqrt{ab}}} \right ) }+{\frac{1}{d}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.78256, size = 992, normalized size = 9.19 \begin{align*} \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 ] \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.63043, size = 204, normalized size = 1.89 \begin{align*} -\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}\left (b\right ) + \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} \end{align*}

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