3.469 \(\int \frac{\sec ^6(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=104 \[ -\frac{\left (3 a^2-2 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac{(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\tan (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.135615, antiderivative size = 104, normalized size of antiderivative = 1., 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} \[ -\frac{\left (3 a^2-2 a b-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac{(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac{\tan (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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 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 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 ^6(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 )^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{b^2}-\frac{a^2-b^2+2 (a-b) b x^2}{b^2 \left (a+b x^2\right )^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\tan (c+d x)}{b^2 d}-\frac{\operatorname{Subst}\left (\int \frac{a^2-b^2+2 (a-b) b x^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{b^2 d}\\ &=\frac{\tan (c+d x)}{b^2 d}+\frac{(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}-\frac{((a-b) (3 a+b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a b^2 d}\\ &=-\frac{(a-b) (3 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{5/2} d}+\frac{\tan (c+d x)}{b^2 d}+\frac{(a-b)^2 \tan (c+d x)}{2 a b^2 d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.620238, size = 104, normalized size = 1. \[ \frac{-\frac{(3 a+b) (a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{a^{3/2}}+\frac{\sqrt{b} (a-b)^2 \sin (2 (c+d x))}{a ((a-b) \cos (2 (c+d x))+a+b)}+2 \sqrt{b} \tan (c+d x)}{2 b^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.085, size = 181, normalized size = 1.7 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{{b}^{2}d}}+{\frac{a\tan \left ( dx+c \right ) }{2\,{b}^{2}d \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{\tan \left ( dx+c \right ) }{db \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{\tan \left ( dx+c \right ) }{2\,ad \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{3\,a}{2\,{b}^{2}d}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{1}{db}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{1}{2\,ad}\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)^6/(a+b*tan(d*x+c)^2)^2,x)

[Out]

tan(d*x+c)/b^2/d+1/2/d/b^2*a*tan(d*x+c)/(a+b*tan(d*x+c)^2)-1/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)-3/2/d/b^2*a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))+1/d/b/(a*b)^(1/2)*arctan(b*tan
(d*x+c)/(a*b)^(1/2))+1/2/d/a/(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)^6/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.72883, size = 1089, normalized size = 10.47 \begin{align*} \left [\frac{{\left ({\left (3 \, a^{3} - 5 \, a^{2} b + a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )\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^{2} +{\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \,{\left (a^{2} b^{4} d \cos \left (d x + c\right ) +{\left (a^{3} b^{3} - a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{3}\right )}}, \frac{{\left ({\left (3 \, a^{3} - 5 \, a^{2} b + a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )\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^{2} +{\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left (a^{2} b^{4} d \cos \left (d x + c\right ) +{\left (a^{3} b^{3} - a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/8*(((3*a^3 - 5*a^2*b + a*b^2 + b^3)*cos(d*x + c)^3 + (3*a^2*b - 2*a*b^2 - b^3)*cos(d*x + c))*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^2 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^4*d*cos(d*x + c) + (a^3*
b^3 - a^2*b^4)*d*cos(d*x + c)^3), 1/4*(((3*a^3 - 5*a^2*b + a*b^2 + b^3)*cos(d*x + c)^3 + (3*a^2*b - 2*a*b^2 -
b^3)*cos(d*x + c))*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^2 + (3*a^3*b - 4*a^2*b^2 + a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^4*d*cos(d*x + c) + (a^3*
b^3 - a^2*b^4)*d*cos(d*x + c)^3)]

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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)**6/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.73124, size = 173, normalized size = 1.66 \begin{align*} \frac{\frac{2 \, \tan \left (d x + c\right )}{b^{2}} - \frac{{\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 )}{\left (3 \, a^{2} - 2 \, a b - b^{2}\right )}}{\sqrt{a b} a b^{2}} + \frac{a^{2} \tan \left (d x + c\right ) - 2 \, a b \tan \left (d x + c\right ) + b^{2} \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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