3.464 \(\int \frac{\sec ^3(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=79 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d \sqrt{a-b}}+\frac{\sin (c+d x)}{2 a d \left (a-(a-b) \sin ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.0795542, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3676, 199, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d \sqrt{a-b}}+\frac{\sin (c+d x)}{2 a d \left (a-(a-b) \sin ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.237823, size = 75, normalized size = 0.95 \[ \frac{\frac{\sqrt{a} \sin (c+d x)}{(b-a) \sin ^2(c+d x)+a}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{\sqrt{a-b}}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.09, size = 80, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{\frac{\sin \left ( dx+c \right ) }{2\,a \left ( a \left ( \sin \left ( dx+c \right ) \right ) ^{2}-b \left ( \sin \left ( dx+c \right ) \right ) ^{2}-a \right ) }}+{\frac{1}{2\,a}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54069, size = 598, normalized size = 7.57 \begin{align*} \left [\frac{{\left ({\left (a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \sqrt{a^{2} - a b} \log \left (-\frac{{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) + 2 \,{\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{3} b - a^{2} b^{2}\right )} d\right )}}, -\frac{{\left ({\left (a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \sqrt{-a^{2} + a b} \arctan \left (\frac{\sqrt{-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) -{\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{3} b - a^{2} b^{2}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(((a - b)*cos(d*x + c)^2 + b)*sqrt(a^2 - a*b)*log(-((a - b)*cos(d*x + c)^2 - 2*sqrt(a^2 - a*b)*sin(d*x +
c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) + 2*(a^2 - a*b)*sin(d*x + c))/((a^4 - 2*a^3*b + a^2*b^2)*d*cos(d*x
+ c)^2 + (a^3*b - a^2*b^2)*d), -1/2*(((a - b)*cos(d*x + c)^2 + b)*sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*si
n(d*x + c)/a) - (a^2 - a*b)*sin(d*x + c))/((a^4 - 2*a^3*b + a^2*b^2)*d*cos(d*x + c)^2 + (a^3*b - a^2*b^2)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral(sec(c + d*x)**3/(a + b*tan(c + d*x)**2)**2, x)

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Giac [A]  time = 1.73869, size = 123, normalized size = 1.56 \begin{align*} \frac{\frac{\arctan \left (-\frac{a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b} a} - \frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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