3.279 \(\int \frac{1}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=77 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac{b \tan (e+f x)}{a f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]
[Out]
ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(3/2)*f) - (b*Tan[e + f*x])/(a*(a + b)*f*Sqrt
[a + b + b*Tan[e + f*x]^2])
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Rubi [A] time = 0.0486906, antiderivative size = 77, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{4128, 382, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac{b \tan (e+f x)}{a f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sec[e + f*x]^2)^(-3/2),x]
[Out]
ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(3/2)*f) - (b*Tan[e + f*x])/(a*(a + b)*f*Sqrt
[a + b + b*Tan[e + f*x]^2])
Rule 4128
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]
Rule 382
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1]) && NeQ[p, -1]
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \tan (e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac{b \tan (e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{a f}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac{b \tan (e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [B] time = 1.46327, size = 168, normalized size = 2.18 \[ \frac{\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt{a+b} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right ) (a \cos (2 (e+f x))+a+2 b)-\sqrt{2} \sqrt{a} b \sin (e+f x) \sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{a+b}}\right )}{4 a^{3/2} f (a+b) \sqrt{\frac{-a \sin ^2(e+f x)+a+b}{a+b}} \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sec[e + f*x]^2)^(-3/2),x]
[Out]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*(Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*(a + 2*
b + a*Cos[2*(e + f*x)]) - Sqrt[2]*Sqrt[a]*b*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)]*Sin[e + f*x]))/(4*a^(
3/2)*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)*Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)])
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Maple [C] time = 0.401, size = 1007, normalized size = 13.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*sec(f*x+e)^2)^(3/2),x)
[Out]
-1/f/(a+b)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/a*(b+a*cos(f*x+e)^2)*(2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2
)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^
(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b)
)^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*sin(f*x+e)+2^(1
/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(
I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))
*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/
(a+b)^2)^(1/2))*b*sin(f*x+e)-2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b
)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e
)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2
)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*sin(f*x+e)-2*
2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+
b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*
x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(
1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b*sin(f*x+e)+cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)*b-((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b)*sin(f*x+e)/(-1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/
cos(f*x+e)^2)^(3/2)/cos(f*x+e)^3
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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(1/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.05488, size = 1438, normalized size = 18.68 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/8*(8*a*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) + ((a^2 + a*b)*cos(f*x + e)
^2 + a*b + b^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b
+ 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a
*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*
b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f
*x + e)^2)*sin(f*x + e)))/((a^4 + a^3*b)*f*cos(f*x + e)^2 + (a^3*b + a^2*b^2)*f), -1/4*(4*a*b*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) + ((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)*sqrt(a)*arcta
n(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a
*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*
sin(f*x + e))))/((a^4 + a^3*b)*f*cos(f*x + e)^2 + (a^3*b + a^2*b^2)*f)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e)**2)**(3/2),x)
[Out]
Integral((a + b*sec(e + f*x)**2)**(-3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e)^2 + a)^(-3/2), x)