3.589 \(\int \frac{(a+b \tan (e+f x))^2}{\sqrt{d \sec (e+f x)}} \, dx\)
Optimal. Leaf size=95 \[ \frac{2 \left (a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{6 a b}{f \sqrt{d \sec (e+f x)}}+\frac{2 b (a+b \tan (e+f x))}{f \sqrt{d \sec (e+f x)}} \]
[Out]
(-6*a*b)/(f*Sqrt[d*Sec[e + f*x]]) + (2*(a^2 - 2*b^2)*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*S
ec[e + f*x]]) + (2*b*(a + b*Tan[e + f*x]))/(f*Sqrt[d*Sec[e + f*x]])
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Rubi [A] time = 0.125565, antiderivative size = 95, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{3508, 3486, 3771, 2639} \[ \frac{2 \left (a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{6 a b}{f \sqrt{d \sec (e+f x)}}+\frac{2 b (a+b \tan (e+f x))}{f \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[e + f*x])^2/Sqrt[d*Sec[e + f*x]],x]
[Out]
(-6*a*b)/(f*Sqrt[d*Sec[e + f*x]]) + (2*(a^2 - 2*b^2)*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*S
ec[e + f*x]]) + (2*b*(a + b*Tan[e + f*x]))/(f*Sqrt[d*Sec[e + f*x]])
Rule 3508
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]
Rule 3486
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])
Rule 3771
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^2}{\sqrt{d \sec (e+f x)}} \, dx &=\frac{2 b (a+b \tan (e+f x))}{f \sqrt{d \sec (e+f x)}}+2 \int \frac{\frac{a^2}{2}-b^2+\frac{3}{2} a b \tan (e+f x)}{\sqrt{d \sec (e+f x)}} \, dx\\ &=-\frac{6 a b}{f \sqrt{d \sec (e+f x)}}+\frac{2 b (a+b \tan (e+f x))}{f \sqrt{d \sec (e+f x)}}+\left (a^2-2 b^2\right ) \int \frac{1}{\sqrt{d \sec (e+f x)}} \, dx\\ &=-\frac{6 a b}{f \sqrt{d \sec (e+f x)}}+\frac{2 b (a+b \tan (e+f x))}{f \sqrt{d \sec (e+f x)}}+\frac{\left (a^2-2 b^2\right ) \int \sqrt{\cos (e+f x)} \, dx}{\sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ &=-\frac{6 a b}{f \sqrt{d \sec (e+f x)}}+\frac{2 \left (a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 b (a+b \tan (e+f x))}{f \sqrt{d \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.885828, size = 64, normalized size = 0.67 \[ \frac{\frac{2 \left (a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\sqrt{\cos (e+f x)}}+2 b (b \tan (e+f x)-2 a)}{f \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[e + f*x])^2/Sqrt[d*Sec[e + f*x]],x]
[Out]
((2*(a^2 - 2*b^2)*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] + 2*b*(-2*a + b*Tan[e + f*x]))/(f*Sqrt[d*Sec[e
+ f*x]])
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Maple [C] time = 0.309, size = 2564, normalized size = 27. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/2),x)
[Out]
1/f*(cos(f*x+e)-1)*(16*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)
+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^3*sin(f*x+e)*b^2+4*(-cos(f*x+e)/(cos(f*x+e)+1)^
2)^(3/2)*cos(f*x+e)^2*a^2-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)^5*a^2+2*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(3/2)*cos(f*x+e)^5*b^2-4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)^4*a^2+2*(-cos(f*x+e)/(cos(f*x+e)
+1)^2)^(3/2)*cos(f*x+e)^4*b^2-4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)^2*b^2-4*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(3/2)*cos(f*x+e)^3*b^2+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)*b^2+2*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(3/2)*cos(f*x+e)*a^2+12*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(
cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^2*sin(f*x+e)*a^2-24*I*(-cos(f*x+e)/(c
os(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/si
n(f*x+e),I)*cos(f*x+e)^2*sin(f*x+e)*b^2-12*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^2*sin(f*x+e)*a^2+24*I*(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*
x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^2*sin(f*x+e)*b^2+8*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^
(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*a^2-16*
I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*
(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*b^2+4*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)
+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*b^2*sin(f*x+e)+2*I*(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*
x+e)-1)/sin(f*x+e),I)*a^2*sin(f*x+e)-4*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*
x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*b^2*sin(f*x+e)-2*I*(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e)
,I)*a^2*sin(f*x+e)-8*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1
))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*a^2+16*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*co
s(f*x+e)*sin(f*x+e)*b^2+2*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x
+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^4*sin(f*x+e)*a^2-4*I*(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e)
,I)*cos(f*x+e)^4*sin(f*x+e)*b^2-2*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/
(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^4*sin(f*x+e)*a^2+4*I*(-cos(f*x+e)/(c
os(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/si
n(f*x+e),I)*cos(f*x+e)^4*sin(f*x+e)*b^2+8*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos
(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^3*sin(f*x+e)*a^2-16*I*(-cos(
f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x
+e)-1)/sin(f*x+e),I)*cos(f*x+e)^3*sin(f*x+e)*b^2-8*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(
1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^3*sin(f*x+e)*a^2+2*
(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^2-a*b*ln(-2*(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*
x+e)^2-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*cos(f*x+e)^2*sin(f*x+e)+a*b*cos(f*
x+e)^2*ln(-(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^
(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*sin(f*x+e)-12*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)^2*sin(f*x+e)
*a*b-4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)*sin(f*x+e)*a*b-4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*c
os(f*x+e)^4*sin(f*x+e)*a*b-12*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*cos(f*x+e)^3*sin(f*x+e)*a*b)*(cos(f*x+e)+1)
^4*(d/cos(f*x+e))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)/cos(f*x+e)^3/d/sin(f*x+e)^3
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\sqrt{d \sec \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e) + a)^2/sqrt(d*sec(f*x + e)), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \sqrt{d \sec \left (f x + e\right )}}{d \sec \left (f x + e\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*sqrt(d*sec(f*x + e))/(d*sec(f*x + e)), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}{\sqrt{d \sec{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))**2/(d*sec(f*x+e))**(1/2),x)
[Out]
Integral((a + b*tan(e + f*x))**2/sqrt(d*sec(e + f*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\sqrt{d \sec \left (f x + e\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e) + a)^2/sqrt(d*sec(f*x + e)), x)