3.470 \(\int \frac {\sec ^4(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=77 \[ \frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac {(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3675, 385, 205} \[ \frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac {(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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Mathematica [A]  time = 0.31, size = 83, normalized size = 1.08 \[ \frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )+\frac {\sqrt {a} \sqrt {b} (b-a) \sin (2 (c+d x))}{(a-b) \cos (2 (c+d x))+a+b}}{2 a^{3/2} b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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fricas [B]  time = 0.55, size = 367, normalized size = 4.77 \[ \left [-\frac {4 \, {\left (a^{2} b - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a b + b^{2}\right )} \sqrt {-a b} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{8 \, {\left (a^{2} b^{3} d + {\left (a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (a^{2} b - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a b + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left (a^{2} b^{3} d + {\left (a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.09, size = 92, normalized size = 1.19 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (a + b\right )}}{\sqrt {a b} a b} - \frac {a \tan \left (d x + c\right ) - b \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.68, size = 112, normalized size = 1.45 \[ -\frac {\tan \left (d x +c \right )}{2 d b \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {\tan \left (d x +c \right )}{2 a d \left (a +b \left (\tan ^{2}\left (d x +c \right )\right )\right )}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d b \sqrt {a b}}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d a \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x)

[Out]

-1/2/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)+1/2/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 = 0.57, size = 69, normalized size = 0.90 \[ -\frac {\frac {{\left (a - b\right )} \tan \left (d x + c\right )}{a b^{2} \tan \left (d x + c\right )^{2} + a^{2} b} - \frac {{\left (a + b\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a b}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.19, size = 65, normalized size = 0.84 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {a}}\right )\,\left (a+b\right )}{2\,a^{3/2}\,b^{3/2}\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a-b\right )}{2\,a\,b\,d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + b*tan(c + d*x)^2)^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*tan(c + d*x)**2)**2, x)

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