3.375 \(\int \frac{\cos (c+d x) (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=348 \[ \frac{A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{a d}+\frac{\sqrt{a+b} (A b-2 a B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^2 d}+\frac{A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{a d}+\frac{A (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a b d} \]

[Out]

(A*(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)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*b*d) + (A*Sqrt[a + b]*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))])/(a*d) + (Sqrt[a + b]*(A*b - 2*a*B)*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)]*Sqrt
[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a^2*d) + (A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(a*d)

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Rubi [A]  time = 0.404588, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4034, 4059, 3921, 3784, 3832, 4004} \[ \frac{\sqrt{a+b} (A b-2 a B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^2 d}+\frac{A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{a d}+\frac{A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a d}+\frac{A (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(A*(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)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*b*d) + (A*Sqrt[a + b]*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))])/(a*d) + (Sqrt[a + b]*(A*b - 2*a*B)*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)]*Sqrt
[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a^2*d) + (A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(a*d)

Rule 4034

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 + 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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4059

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A
- 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, 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 \frac{\cos (c+d x) (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx &=\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}-\frac{\int \frac{\frac{1}{2} (A b-2 a B)+\frac{1}{2} A b \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{a}\\ &=\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}-\frac{\int \frac{\frac{1}{2} (A b-2 a B)-\frac{1}{2} A b \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{a}-\frac{(A b) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 a}\\ &=\frac{A (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}}}{a b d}+\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}+\frac{(A b) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 a}-\frac{(A b-2 a B) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 a}\\ &=\frac{A (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}}}{a b d}+\frac{A \sqrt{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}}}{a d}+\frac{\sqrt{a+b} (A b-2 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^2 d}+\frac{A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{a d}\\ \end{align*}

Mathematica [C]  time = 17.0238, size = 1027, normalized size = 2.95 \[ \frac{\sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)} \sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \left (-a A \sqrt{\frac{b-a}{a+b}} \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \tan ^3\left (\frac{1}{2} (c+d x)\right )+A b \sqrt{\frac{b-a}{a+b}} \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \tan ^3\left (\frac{1}{2} (c+d x)\right )+2 i A 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{\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 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{\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 A \sqrt{\frac{b-a}{a+b}} \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \tan \left (\frac{1}{2} (c+d x)\right )+A b \sqrt{\frac{b-a}{a+b}} \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \tan \left (\frac{1}{2} (c+d x)\right )+2 i A 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{\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 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{\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}}-i A (a-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 ) \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-a B) \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 ) \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}}\right )}{a \sqrt{\frac{b-a}{a+b}} d \sqrt{a+b \sec (c+d x)} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^{3/2} \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}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*
(a*A*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]*Sqrt[1 - Tan[(c + d*x)/2]^2] + A*b*Sqrt[(-a + b)/(a + b)]*Tan[(c
+ d*x)/2]*Sqrt[1 - Tan[(c + d*x)/2]^2] - a*A*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3*Sqrt[1 - Tan[(c + d*x)/
2]^2] + A*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3*Sqrt[1 - Tan[(c + d*x)/2]^2] + (2*I)*A*b*EllipticPi[-((a
+ b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[(a + b - a*Tan[(c +
d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (4*I)*a*B*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/
(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b
)] + (2*I)*A*b*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[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (4*I)*a*B*Ellipt
icPi[-((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[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - I*A*(a - b)*EllipticE[I*ArcSinh[Sqrt[
(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*(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 - a*B)*EllipticF[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d
*x)/2]], (a + b)/(a - b)]*(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)]))/(a*Sqrt[(-a + b)/(a + b)]*d*Sqrt[a + b*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 = 0.39, size = 1028, normalized size = 3. \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+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x)

[Out]

1/d/a*(-1+cos(d*x+c))^2*(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))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b-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))/s
in(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-4*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)*EllipticPi((-1+cos(d*x+
c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a+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)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a+2*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+
c),-1,((a-b)/(a+b))^(1/2))*b*sin(d*x+c)-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)/(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*sin(d*x+c)-4*B*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)*a+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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(
1/2))*a-A*cos(d*x+c)^3*a+A*a*cos(d*x+c)^2-A*cos(d*x+c)^2*b+A*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 \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\sqrt{b \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\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+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))/sqrt(b*sec(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x))*cos(c + d*x)/sqrt(a + b*sec(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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