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

Optimal. Leaf size=40 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b+c} \tan (c+d x)}{\sqrt{a+c}}\right )}{d \sqrt{a+c} \sqrt{b+c}} \]

[Out]

ArcTan[(Sqrt[b + c]*Tan[c + d*x])/Sqrt[a + c]]/(Sqrt[a + c]*Sqrt[b + c]*d)

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Rubi [A]  time = 0.605453, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.03, Rules used = {205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b+c} \tan (c+d x)}{\sqrt{a+c}}\right )}{d \sqrt{a+c} \sqrt{b+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[b + c]*Tan[c + d*x])/Sqrt[a + c]]/(Sqrt[a + c]*Sqrt[b + c]*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]

Rubi steps

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

Mathematica [A]  time = 0.263189, size = 40, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b+c} \tan (c+d x)}{\sqrt{a+c}}\right )}{d \sqrt{a+c} \sqrt{b+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[b + c]*Tan[c + d*x])/Sqrt[a + c]]/(Sqrt[a + c]*Sqrt[b + c]*d)

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Maple [A]  time = 0.084, size = 34, normalized size = 0.9 \begin{align*}{\frac{1}{d}\arctan \left ({ \left ( b+c \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( a+c \right ) \left ( b+c \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+c \right ) \left ( b+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.93774, size = 747, normalized size = 18.68 \begin{align*} \left [-\frac{\sqrt{-a b -{\left (a + b\right )} c - c^{2}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2} + 8 \,{\left (a + b\right )} c + 8 \, c^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, a b + b^{2} +{\left (3 \, a + 5 \, b\right )} c + 4 \, c^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left ({\left (a + b + 2 \, c\right )} \cos \left (d x + c\right )^{3} -{\left (b + c\right )} \cos \left (d x + c\right )\right )} \sqrt{-a b -{\left (a + b\right )} c - c^{2}} \sin \left (d x + c\right ) + b^{2} + 2 \, b c + c^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (a b - b^{2} +{\left (a - b\right )} c\right )} \cos \left (d x + c\right )^{2} + b^{2} + 2 \, b c + c^{2}}\right )}{4 \,{\left (a b +{\left (a + b\right )} c + c^{2}\right )} d}, -\frac{\arctan \left (\frac{{\left (a + b + 2 \, c\right )} \cos \left (d x + c\right )^{2} - b - c}{2 \, \sqrt{a b +{\left (a + b\right )} c + c^{2}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, \sqrt{a b +{\left (a + b\right )} c + c^{2}} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*sqrt(-a*b - (a + b)*c - c^2)*log(((a^2 + 6*a*b + b^2 + 8*(a + b)*c + 8*c^2)*cos(d*x + c)^4 - 2*(3*a*b +
b^2 + (3*a + 5*b)*c + 4*c^2)*cos(d*x + c)^2 + 4*((a + b + 2*c)*cos(d*x + c)^3 - (b + c)*cos(d*x + c))*sqrt(-a*
b - (a + b)*c - c^2)*sin(d*x + c) + b^2 + 2*b*c + c^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2 + (a
- b)*c)*cos(d*x + c)^2 + b^2 + 2*b*c + c^2))/((a*b + (a + b)*c + c^2)*d), -1/2*arctan(1/2*((a + b + 2*c)*cos(
d*x + c)^2 - b - c)/(sqrt(a*b + (a + b)*c + c^2)*cos(d*x + c)*sin(d*x + c)))/(sqrt(a*b + (a + b)*c + c^2)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.42797, size = 103, normalized size = 2.58 \begin{align*} \frac{\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, b + 2 \, c\right ) + \arctan \left (\frac{b \tan \left (d x + c\right ) + c \tan \left (d x + c\right )}{\sqrt{a b + a c + b c + c^{2}}}\right )}{\sqrt{a b + a c + b c + c^{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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