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

Optimal. Leaf size=352 \[ \frac{\sqrt{a+b} (2 C (a-b)+A b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{b d}+\frac{(a-b) \sqrt{a+b} (A-2 C) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b d}+\frac{A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{d}-\frac{A b \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d} \]

[Out]

((a - b)*Sqrt[a + b]*(A - 2*C)*Cot[c + d*x]*EllipticE[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))])/(b*d) + (Sqrt[a + b]*(A*b
+ 2*(a - b)*C)*Cot[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))])/(b*d) - (A*b*Sqrt[a + b]*Cot[c + d*x]*El
lipticPi[(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))])/(a*d) + (A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d

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Rubi [A]  time = 0.379191, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4095, 4058, 3921, 3784, 3832, 4004} \[ \frac{\sqrt{a+b} (2 C (a-b)+A b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b d}+\frac{(a-b) \sqrt{a+b} (A-2 C) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b d}+\frac{A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{d}-\frac{A b \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*Sqrt[a + b]*(A - 2*C)*Cot[c + d*x]*EllipticE[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))])/(b*d) + (Sqrt[a + b]*(A*b
+ 2*(a - b)*C)*Cot[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))])/(b*d) - (A*b*Sqrt[a + b]*Cot[c + d*x]*El
lipticPi[(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))])/(a*d) + (A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d

Rule 4095

Int[((A_.) + 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*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e +
f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 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 (c+d x) \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}+\int \frac{\frac{A b}{2}+a C \sec (c+d x)-\frac{1}{2} b (A-2 C) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}-\frac{1}{2} (b (A-2 C)) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx+\int \frac{\frac{A b}{2}+\left (\frac{1}{2} b (A-2 C)+a C\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{(a-b) \sqrt{a+b} (A-2 C) \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{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{2} (A b) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{2} (A b+2 (a-b) C) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{(a-b) \sqrt{a+b} (A-2 C) \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} (A b+2 (a-b) 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}}}{b d}-\frac{A b \sqrt{a+b} \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{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [B]  time = 18.2512, size = 727, normalized size = 2.07 \[ \frac{\sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \sqrt{a+b \sec (c+d x)} \left (-2 (A b-C (a+b)) \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 )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{a+b}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+(a+b) (A-2 C) \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 )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )-2 A b \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \tan ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{a+b}} \Pi \left (-1;-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )-2 A b \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 )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{a+b}} \Pi \left (-1;-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )+a A \tan ^5\left (\frac{1}{2} (c+d x)\right )-2 a A \tan ^3\left (\frac{1}{2} (c+d x)\right )+a A \tan \left (\frac{1}{2} (c+d x)\right )-2 a C \tan ^5\left (\frac{1}{2} (c+d x)\right )+4 a C \tan ^3\left (\frac{1}{2} (c+d x)\right )-2 a C \tan \left (\frac{1}{2} (c+d x)\right )-A b \tan ^5\left (\frac{1}{2} (c+d x)\right )+A b \tan \left (\frac{1}{2} (c+d x)\right )+2 b C \tan ^5\left (\frac{1}{2} (c+d x)\right )-2 b C \tan \left (\frac{1}{2} (c+d x)\right )\right )}{d \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+b} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+b}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}}}+\frac{2 C \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.508, size = 1602, normalized size = 4.6 \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)*(A+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)

[Out]

-1/d*(-1+cos(d*x+c))^2*(A*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))/(c
os(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a+A*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))/sin(d*x+
c),((a-b)/(a+b))^(1/2))*b-2*A*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))*b+2*A*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))/s
in(d*x+c),-1,((a-b)/(a+b))^(1/2))*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a-2*C*sin(d*x+c)*c
os(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*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*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))*b+A*(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*sin(d*x+c)+A*(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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)
/(a+b))^(1/2))*b*sin(d*x+c)-2*A*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*A*EllipticPi((-1+cos(d*x+c))/sin(
d*x+c),-1,((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*C*EllipticE((-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*EllipticE((-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*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+A*cos(d*x+c)^3*a
-A*cos(d*x+c)^2*a+A*cos(d*x+c)^2*b+2*C*cos(d*x+c)^2*a-A*cos(d*x+c)*b-2*C*cos(d*x+c)*a+2*C*cos(d*x+c)*b-2*C*b)*
(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} + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+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 + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c), 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 ) \sec \left (d x + c\right )^{2} + A \cos \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)*(A+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)*sec(d*x + c)^2 + A*cos(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)*(A+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} + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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