3.536 \(\int \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=396 \[ \frac{\sqrt{a+b} (2 a+b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{4 a d}-\frac{\sqrt{a+b} \left (4 a^2-b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^2 d}+\frac{b \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{4 a d}+\frac{\sin (c+d x) \cos (c+d x) \sqrt{a+b \sec (c+d x)}}{2 d}+\frac{(a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a d} \]

[Out]

((a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqr
t[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a*d) + (Sqrt[a + b]*(2*a + b)*Co
t[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a*d) - (Sqrt[a + b]*(4*a^2 - b^2)*Cot[c + d*x]*Elliptic
Pi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a +
b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a^2*d) + (b*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(4*a*d) + (
Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.598746, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3857, 4104, 4058, 3921, 3784, 3832, 4004} \[ -\frac{\sqrt{a+b} \left (4 a^2-b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a^2 d}+\frac{b \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{4 a d}+\frac{\sin (c+d x) \cos (c+d x) \sqrt{a+b \sec (c+d x)}}{2 d}+\frac{\sqrt{a+b} (2 a+b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a d}+\frac{(a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]],x]

[Out]

((a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqr
t[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a*d) + (Sqrt[a + b]*(2*a + b)*Co
t[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a*d) - (Sqrt[a + b]*(4*a^2 - b^2)*Cot[c + d*x]*Elliptic
Pi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a +
b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(4*a^2*d) + (b*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(4*a*d) + (
Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*d)

Rule 3857

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(Cot[e
+ f*x]*Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n + 1)*
Simp[b - 2*a*(n + 1)*Csc[e + f*x] - b*(2*n + 3)*Csc[e + f*x]^2, x])/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[
{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4058

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, 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, C}, x] && NeQ[a^2 - b^2, 0
]

Rule 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], 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 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 \cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \, dx &=\frac{\cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{1}{4} \int \frac{\cos (c+d x) \left (b+2 a \sec (c+d x)+b \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac{\cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{\int \frac{\frac{1}{2} \left (-4 a^2+b^2\right )-a b \sec (c+d x)+\frac{1}{2} b^2 \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{4 a}\\ &=\frac{b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac{\cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{2 d}-\frac{\int \frac{\frac{1}{2} \left (-4 a^2+b^2\right )+\left (-a b-\frac{b^2}{2}\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{4 a}-\frac{b^2 \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{8 a}\\ &=\frac{(a-b) \sqrt{a+b} \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac{b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac{\cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{2 d}+\frac{(b (2 a+b)) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{8 a}+\frac{\left (4 a^2-b^2\right ) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{8 a}\\ &=\frac{(a-b) \sqrt{a+b} \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a d}+\frac{\sqrt{a+b} (2 a+b) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a d}-\frac{\sqrt{a+b} \left (4 a^2-b^2\right ) \cot (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{4 a^2 d}+\frac{b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 a d}+\frac{\cos (c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [C]  time = 18.5592, size = 1173, normalized size = 2.96 \[ \frac{\sqrt{a+b \sec (c+d x)} \sin (2 (c+d x))}{4 d}+\frac{\sqrt{a+b \sec (c+d x)} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}} \left (-b^2 \sqrt{\frac{b-a}{a+b}} \tan ^5\left (\frac{1}{2} (c+d x)\right )+a b \sqrt{\frac{b-a}{a+b}} \tan ^5\left (\frac{1}{2} (c+d x)\right )-2 a b \sqrt{\frac{b-a}{a+b}} \tan ^3\left (\frac{1}{2} (c+d x)\right )-8 i a^2 \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}} \tan ^2\left (\frac{1}{2} (c+d x)\right )+2 i b^2 \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}} \tan ^2\left (\frac{1}{2} (c+d x)\right )+b^2 \sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )+a b \sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )-i (a-b) b E\left (i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right ) \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}}+2 i \left (2 a^2-b a-b^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right ) \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}}-8 i a^2 \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}}+2 i b^2 \Pi \left (-\frac{a+b}{a-b};i \sinh ^{-1}\left (\sqrt{\frac{b-a}{a+b}} \tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a+b}{a-b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}}\right )}{4 a \sqrt{\frac{b-a}{a+b}} d \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) \sqrt{\frac{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \left (a \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )-b \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(Sqrt[a + b*Sec[c + d*x]]*Sin[2*(c + d*x)])/(4*d) + (Sqrt[a + b*Sec[c + d*x]]*Sqrt[(a + b - a*Tan[(c + d*x)/2]
^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(a*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + b^2*Sqrt[(
-a + b)/(a + b)]*Tan[(c + d*x)/2] - 2*a*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 + a*b*Sqrt[(-a + b)/(a + b
)]*Tan[(c + d*x)/2]^5 - b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - (8*I)*a^2*EllipticPi[-((a + b)/(a - b)
), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a +
b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (2*I)*b^2*EllipticPi[-((a + b)/(a - b)), I*ArcSin
h[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[
(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (8*I)*a^2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a
+ b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b
- a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (2*I)*b^2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[S
qrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt
[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - I*(a - b)*b*EllipticE[I*ArcSinh[Sqrt[(-a + b
)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a +
b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (2*I)*(2*a^2 - a*b - b^2)*EllipticF[I*ArcSinh[Sqr
t[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*
Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(4*a*Sqrt[(-a + b)/(a + b)]*d*Sqrt[b + a
*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2
]^2)]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))

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Maple [B]  time = 0.286, size = 1257, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x)

[Out]

1/4/d/a*(-1+cos(d*x+c))^2*(4*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2-2*cos(d*x+c)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a
-b)/(a+b))^(1/2))*sin(d*x+c)*a*b-cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d
*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b-sin(d*x+c)*cos(d*x+c)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*b^2-8*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos
(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^2+2*cos(d*x+c)*
b^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*EllipticPi((-
1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))+4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*sin(d*x+c)-2*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a
-b)/(a+b))^(1/2))*a*b*sin(d*x+c)-(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(
1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b*sin(d*x+c)-(cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^
2*sin(d*x+c)-8*a^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticP
i((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)+2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b^2*sin(d*
x+c)-2*cos(d*x+c)^4*a^2-3*cos(d*x+c)^3*a*b+2*cos(d*x+c)^2*a^2+cos(d*x+c)^2*a*b-cos(d*x+c)^2*b^2+2*cos(d*x+c)*a
*b+cos(d*x+c)*b^2)*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(d*x + c) + a)*cos(d*x + c)^2, x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a + b*sec(c + d*x))*cos(c + d*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(d*x + c) + a)*cos(d*x + c)^2, x)