3.6.89 \(\int \frac {(a+b \tan (e+f x))^2}{\sqrt {d \sec (e+f x)}} \, dx\) [589]
Optimal. Leaf size=95 \[ -\frac {6 a b}{f \sqrt {d \sec (e+f x)}}+\frac {2 \left (a^2-2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 b (a+b \tan (e+f x))}{f \sqrt {d \sec (e+f x)}} \]
[Out]
-6*a*b/f/(d*sec(f*x+e))^(1/2)+2*(a^2-2*b^2)*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*
f*x+1/2*e),2^(1/2))/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)+2*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(1/2)
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Rubi [A]
time = 0.10, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3589, 3567,
3856, 2719} \begin {gather*} \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)}} \end {gather*}
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 2719
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]
Rule 3567
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 3589
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[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 3856
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]
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*}
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Mathematica [A]
time = 1.01, size = 64, normalized size = 0.67 \begin {gather*} \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 (-2 a+b \tan (e+f x))}{f \sqrt {d \sec (e+f x)}} \end {gather*}
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] Result contains complex when optimal does not.
time = 0.54, size = 2565, normalized size = 27.00
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\(\text {Expression too large to display}\) |
\(2565\) |
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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,method=_RETURNVERBOSE)
[Out]
-1/f*(cos(f*x+e)-1)*(-8*I*cos(f*x+e)*sin(f*x+e)*(-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^2-2*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)-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)+4*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^2+16*I*cos(f*
x+e)^3*sin(f*x+e)*(-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+8*I*cos(f*x+e)^3*sin(f*x+e)*(-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^2-16
*I*cos(f*x+e)^3*sin(f*x+e)*(-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^2-12*I*cos(f*x+e)^2*sin(f*x+e)*(-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+24*I*cos(f*x+e)^2*sin(f*x+e)*(-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+12*I*cos(f*x+e)^2*sin(f*x+e)*(-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^2-24*I*cos(f*x+e)^2*sin(f*x+e)*(-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+16*I*cos(f*x+e)*sin(f*x+e)*(-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+8*I*cos(f*x+e)*sin(f*x+e)*(-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^2-16*I*cos(f*x+e)*sin(f*x+e)*
(-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*(c
os(f*x+e)-1)/sin(f*x+e),I)*b^2-2*I*cos(f*x+e)^4*sin(f*x+e)*(-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+4*I*cos(f*x+e)^4*sin
(f*x+e)*(-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)*Ellipt
icF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*b^2+2*I*cos(f*x+e)^4*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(co
s(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^2-4*I*cos(f*x+
e)^4*sin(f*x+e)*(-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^2-8*I*cos(f*x+e)^3*sin(f*x+e)*(-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-a*b*
cos(f*x+e)^2*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)
/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*sin(f*x+e)+a*b*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^
(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*cos(f*x+e)^2*sin(f*x+e
)+12*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a*b-4*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(3/2)*a^2+4*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^2-2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(3/2)*b^2-2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a^2+2*cos(f*x+e)^5*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(3/2)*a^2-2*cos(f*x+e)^5*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^2+4*cos(f*x+e)^4*(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(3/2)*a^2-2*cos(f*x+e)^4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^2-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/
2)*b^2+4*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a*b+4*cos(f*x+e)^4*sin(f*x+e)*(-cos(f*x+e)
/(cos(f*x+e)+1)^2)^(3/2)*a*b+12*cos(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
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 [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.13, size = 129, normalized size = 1.36 \begin {gather*} \frac {\sqrt {2} {\left (i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {2} {\left (-i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (2 \, a b \cos \left (f x + e\right ) - b^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{d f} \end {gather*}
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]
(sqrt(2)*(I*a^2 - 2*I*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x
+ e))) + sqrt(2)*(-I*a^2 + 2*I*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I
*sin(f*x + e))) - 2*(2*a*b*cos(f*x + e) - b^2*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d*f)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \end {gather*}
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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(1/2),x)
[Out]
int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(1/2), x)
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