3.198 \(\int \sqrt{a+b \sec (e+f x)} (c+d \sec (e+f x)) \, dx\)
Optimal. Leaf size=320 \[ \frac{2 \sqrt{a+b} (a d+b (c-d)) \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{b f}-\frac{2 c \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{f}-\frac{2 d (a-b) \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b f} \]
[Out]
(-2*(a - b)*Sqrt[a + b]*d*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)
]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(b*f) + (2*Sqrt[a + b]*(b*(c -
d) + a*d)*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 -
Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(b*f) - (2*Sqrt[a + b]*c*Cot[e + f*x]*Ellipti
cPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a +
b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/f
________________________________________________________________________________________
Rubi [A] time = 0.282933, antiderivative size = 320, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3916, 3784, 4005, 3832, 4004} \[ \frac{2 \sqrt{a+b} (a d+b (c-d)) \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b f}-\frac{2 c \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{f}-\frac{2 d (a-b) \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b f} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x]),x]
[Out]
(-2*(a - b)*Sqrt[a + b]*d*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)
]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(b*f) + (2*Sqrt[a + b]*(b*(c -
d) + a*d)*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 -
Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(b*f) - (2*Sqrt[a + b]*c*Cot[e + f*x]*Ellipti
cPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a +
b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/f
Rule 3916
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[a*c,
Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Int[(Csc[e + f*x]*(b*c + a*d + b*d*Csc[e + f*x]))/Sqrt[a + b*Csc[e +
f*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Rule 3784
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Rule 4005
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4004
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rubi steps
\begin{align*} \int \sqrt{a+b \sec (e+f x)} (c+d \sec (e+f x)) \, dx &=(a c) \int \frac{1}{\sqrt{a+b \sec (e+f x)}} \, dx+\int \frac{\sec (e+f x) (b c+a d+b d \sec (e+f x))}{\sqrt{a+b \sec (e+f x)}} \, dx\\ &=-\frac{2 \sqrt{a+b} c \cot (e+f x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{f}+(b d) \int \frac{\sec (e+f x) (1+\sec (e+f x))}{\sqrt{a+b \sec (e+f x)}} \, dx+(b (c-d)+a d) \int \frac{\sec (e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx\\ &=-\frac{2 (a-b) \sqrt{a+b} d \cot (e+f x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{b f}+\frac{2 \sqrt{a+b} (b (c-d)+a d) \cot (e+f x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{b f}-\frac{2 \sqrt{a+b} c \cot (e+f x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{f}\\ \end{align*}
Mathematica [C] time = 17.8105, size = 913, normalized size = 2.85 \[ \frac{2 d \cos (e+f x) \sqrt{a+b \sec (e+f x)} (c+d \sec (e+f x)) \sin (e+f x)}{f (d+c \cos (e+f x))}+\frac{2 \sqrt{a+b \sec (e+f x)} (c+d \sec (e+f x)) \left (a \sqrt{\frac{b-a}{a+b}} d \tan ^5\left (\frac{1}{2} (e+f x)\right )-b \sqrt{\frac{b-a}{a+b}} d \tan ^5\left (\frac{1}{2} (e+f x)\right )-2 a \sqrt{\frac{b-a}{a+b}} d \tan ^3\left (\frac{1}{2} (e+f x)\right )+2 i a c \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (e+f x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (e+f x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a+b}{a+b}} \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \sqrt{\frac{b-a}{a+b}} d \tan \left (\frac{1}{2} (e+f x)\right )+b \sqrt{\frac{b-a}{a+b}} d \tan \left (\frac{1}{2} (e+f x)\right )-i (a-b) d E\left (i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (e+f x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (e+f x)\right )} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right ) \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a+b}{a+b}}-i (a-b) (c-d) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (e+f x)\right )\right ),\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (e+f x)\right )} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right ) \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a+b}{a+b}}+2 i a c \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (e+f x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (e+f x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a+b}{a+b}}\right )}{\sqrt{\frac{b-a}{a+b}} f \sqrt{b+a \cos (e+f x)} (d+c \cos (e+f x)) \sec ^{\frac{3}{2}}(e+f x) \sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (e+f x)\right )}} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right ) \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )^{3/2} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a+b}{\tan ^2\left (\frac{1}{2} (e+f x)\right )+1}}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x]),x]
[Out]
(2*d*Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x])*Sin[e + f*x])/(f*(d + c*Cos[e + f*x])) + (2*Sq
rt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x])*(a*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2] + b*Sqrt[(-a + b)/(a
+ b)]*d*Tan[(e + f*x)/2] - 2*a*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2]^3 + a*Sqrt[(-a + b)/(a + b)]*d*Tan[(e
+ f*x)/2]^5 - b*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2]^5 + (2*I)*a*c*EllipticPi[-((a + b)/(a - b)), I*ArcS
inh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Ta
n[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + (2*I)*a*c*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-
a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Tan[(e + f*x)/2]^2*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a +
b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - I*(a - b)*d*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a +
b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(a + b - a
*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - I*(a - b)*(c - d)*EllipticF[I*ArcSinh[Sqrt[(-a + b)/(a
+ b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(a + b -
a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)]))/(Sqrt[(-a + b)/(a + b)]*f*Sqrt[b + a*Cos[e + f*x]]*(d
+ c*Cos[e + f*x])*Sec[e + f*x]^(3/2)*Sqrt[(1 - Tan[(e + f*x)/2]^2)^(-1)]*(-1 + Tan[(e + f*x)/2]^2)*(1 + Tan[(e
+ f*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(1 + Tan[(e + f*x)/2]^2)])
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Maple [B] time = 0.38, size = 1372, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sec(f*x+e))*(a+b*sec(f*x+e))^(1/2),x)
[Out]
2/f*(1/cos(f*x+e)*(a*cos(f*x+e)+b))^(1/2)*(1+cos(f*x+e))^2*(-1+cos(f*x+e))^2*(cos(f*x+e)*(cos(f*x+e)/(1+cos(f*
x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),(
(a-b)/(a+b))^(1/2))*a*c-cos(f*x+e)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))
^(1/2)*sin(f*x+e)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*d-cos(f*x+e)*(cos(f*x+e)/(1+cos(
f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticF((-1+cos(f*x+e))/sin(f*x+e)
,((a-b)/(a+b))^(1/2))*b*c-cos(f*x+e)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)
))^(1/2)*sin(f*x+e)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b*d+cos(f*x+e)*(cos(f*x+e)/(1+co
s(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticE((-1+cos(f*x+e))/sin(f*x+
e),((a-b)/(a+b))^(1/2))*a*d+cos(f*x+e)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+
e)))^(1/2)*sin(f*x+e)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b*d-2*cos(f*x+e)*(cos(f*x+e)/(
1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticPi((-1+cos(f*x+e))/sin
(f*x+e),-1,((a-b)/(a+b))^(1/2))*a*c+(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e))
)^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*c*sin(f*x+e)-(cos(f*x+e)/(1+cos(f*x+e)))^(
1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))
*a*d*sin(f*x+e)-(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-
1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b*c*sin(f*x+e)-(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos
(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b*d*sin(f*x+e)+(cos
(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*EllipticE((-1+cos(f*x
+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*d+(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*
x+e)))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b*d*sin(f*x+e)-2*(cos(f*x+e)/(1+cos(f*x
+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a
+b))^(1/2))*a*c*sin(f*x+e)-cos(f*x+e)^2*a*d+cos(f*x+e)*a*d-cos(f*x+e)*b*d+d*b)/sin(f*x+e)^5/(a*cos(f*x+e)+b)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right ) + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))*(a+b*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sec \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right ) + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))*(a+b*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec{\left (e + f x \right )}} \left (c + d \sec{\left (e + f x \right )}\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))*(a+b*sec(f*x+e))**(1/2),x)
[Out]
Integral(sqrt(a + b*sec(e + f*x))*(c + d*sec(e + f*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right ) + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))*(a+b*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c), x)