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

Optimal. Leaf size=344 \[ \frac{\sqrt{a+b} (B+2 C) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{d}-\frac{\sqrt{a+b} (2 a C+b B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d}+\frac{B \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{d}+\frac{B (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b d} \]

[Out]

((a - b)*Sqrt[a + b]*B*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (Sqrt[a + b]*(B + 2*C)*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))])/d - (Sqrt[a + b]*(b*B + 2*a*C)*Cot[c + d*x]*EllipticPi[(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)]*Sq
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d) + (B*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.46364, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4072, 4032, 4058, 3921, 3784, 3832, 4004} \[ \frac{\sqrt{a+b} (B+2 C) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}-\frac{\sqrt{a+b} (2 a C+b B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d}+\frac{B \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{d}+\frac{B (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*Sqrt[a + b]*B*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) + (Sqrt[a + b]*(B + 2*C)*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))])/d - (Sqrt[a + b]*(b*B + 2*a*C)*Cot[c + d*x]*EllipticPi[(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)]*Sq
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d) + (B*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d

Rule 4072

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

Rule 4032

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*A*(n + 1))*C
sc[e + f*x] - A*b*(m + n + 1)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B,
0] && NeQ[a^2 - b^2, 0] && LtQ[0, m, 1] && 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)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos (c+d x) \sqrt{a+b \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac{B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}+\int \frac{\frac{1}{2} (b B+2 a C)+b C \sec (c+d x)-\frac{1}{2} b B \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}-\frac{1}{2} (b B) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx+\int \frac{\frac{1}{2} (b B+2 a C)+\left (\frac{b B}{2}+b C\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{(a-b) \sqrt{a+b} 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}}}{b d}+\frac{B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{2} (b (B+2 C)) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{2} (b B+2 a C) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{(a-b) \sqrt{a+b} 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}}}{b d}+\frac{\sqrt{a+b} (B+2 C) \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}}}{d}-\frac{\sqrt{a+b} (b B+2 a C) \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}}}{a d}+\frac{B \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [C]  time = 18.0138, size = 1107, normalized size = 3.22 \[ \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 (a \sqrt{\frac{b-a}{a+b}} B \tan ^5\left (\frac{1}{2} (c+d x)\right )-b \sqrt{\frac{b-a}{a+b}} B \tan ^5\left (\frac{1}{2} (c+d x)\right )-2 a \sqrt{\frac{b-a}{a+b}} B \tan ^3\left (\frac{1}{2} (c+d x)\right )-2 i b B \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 )-4 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} (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 )+a \sqrt{\frac{b-a}{a+b}} B \tan \left (\frac{1}{2} (c+d x)\right )+b \sqrt{\frac{b-a}{a+b}} 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 (a-b) C \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}}-2 i b B \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}}-4 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} (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 )}{\sqrt{\frac{b-a}{a+b}} d \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)} \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 (-b \tan ^4\left (\frac{1}{2} (c+d x)\right )+a \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )^2+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2] + b*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2] - 2*a*Sqrt[(-a + b
)/(a + b)]*B*Tan[(c + d*x)/2]^3 + a*Sqrt[(-a + b)/(a + b)]*B*Tan[(c + d*x)/2]^5 - b*Sqrt[(-a + b)/(a + b)]*B*T
an[(c + d*x)/2]^5 - (2*I)*b*B*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)] - (4*I)*a*C*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*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Ta
n[(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)] - (4*I)*a*C*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)] - 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)*(a - b)*C*EllipticF[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)]))/(Sqrt[(-a + b)/(a + b)]*d*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1
- Tan[(c + d*x)/2]^2)]*(b - b*Tan[(c + d*x)/2]^4 + a*(-1 + Tan[(c + d*x)/2]^2)^2))

________________________________________________________________________________________

Maple [B]  time = 0.423, size = 1386, normalized size = 4. \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)

[Out]

-1/d*(-1+cos(d*x+c))^2*(-2*B*cos(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b+2*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1
,((a-b)/(a+b))^(1/2))*sin(d*x+c)*b+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))/(c
os(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*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)*b-2*C*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*C*EllipticF((
-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*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)*sin(d*x+c)*b+4*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))*sin(d*x+c)*a-2*B*E
llipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b+2*B*(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))*b+B*(cos(d*x+c)/(co
s(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)*EllipticE((-1+cos(d*x+c))/sin(d*
x+c),((a-b)/(a+b))^(1/2))*a+B*(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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b-2*C*EllipticF((-1+cos(d*x+c))/sin(d*x
+c),((a-b)/(a+b))^(1/2))*(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)*a+2*C*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b+4*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*c
os(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*sin(d*x+c)+B*
cos(d*x+c)^3*a-B*a*cos(d*x+c)^2+B*cos(d*x+c)^2*b-B*b*cos(d*x+c))*(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{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)

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Sympy [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*(B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+b*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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