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

Optimal. Leaf size=52 \[ \frac {\tan (c+d x)}{b d}-\frac {(a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2} d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 52, 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, 388, 205} \[ \frac {\tan (c+d x)}{b d}-\frac {(a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*b^(3/2)*d)) + Tan[c + d*x]/(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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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)}{a+b \tan ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\tan (c+d x)}{b d}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=-\frac {(a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2} d}+\frac {\tan (c+d x)}{b d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 52, normalized size = 1.00 \[ \frac {\tan (c+d x)}{b d}-\frac {(a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 267, normalized size = 5.13 \[ \left [\frac {\sqrt {-a b} {\left (a - b\right )} \cos \left (d x + c\right ) \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, a b \sin \left (d x + c\right )}{4 \, a b^{2} d \cos \left (d x + c\right )}, \frac {\sqrt {a b} {\left (a - b\right )} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) + 2 \, a b \sin \left (d x + c\right )}{2 \, a b^{2} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.38, size = 62, normalized size = 1.19 \[ -\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} {\left (a - b\right )}}{\sqrt {a b} b} - \frac {\tan \left (d x + c\right )}{b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.59, size = 66, normalized size = 1.27 \[ \frac {\tan \left (d x +c \right )}{b d}-\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right ) a}{d b \sqrt {a b}}+\frac {\arctan \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {a b}}\right )}{d \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(d*x+c)/b/d-1/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 = 0.74, size = 45, normalized size = 0.87 \[ -\frac {\frac {{\left (a - b\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {\tan \left (d x + c\right )}{b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.40, size = 44, normalized size = 0.85 \[ \frac {\mathrm {tan}\left (c+d\,x\right )}{b\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {a}}\right )\,\left (a-b\right )}{\sqrt {a}\,b^{3/2}\,d} \]

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)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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