Integrand size = 33, antiderivative size = 352 \[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} (A-2 C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}+\frac {\sqrt {a+b} (A b+2 (a-b) C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}-\frac {A b \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d}+\frac {A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{d} \] Output:
(a-b)*(a+b)^(1/2)*(A-2*C)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b )^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x +c))/(a-b))^(1/2)/b/d+(a+b)^(1/2)*(A*b+2*(a-b)*C)*cot(d*x+c)*EllipticF((a+ b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+ b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d-A*b*(a+b)^(1/2)*cot(d*x+c)*E llipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))* (b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/d+A*(a+b* sec(d*x+c))^(1/2)*sin(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(723\) vs. \(2(352)=704\).
Time = 16.42 (sec) , antiderivative size = 723, normalized size of antiderivative = 2.05 \[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
Integrate[Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2),x]
Output:
(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 - 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*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*Tan[(c + d *x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(A - 2*C)*EllipticE[Ar cSin[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 - 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)]))/(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)])
Time = 1.25 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4583, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4583 |
\(\displaystyle \int \frac {-b (A-2 C) \sec ^2(c+d x)+2 a C \sec (c+d x)+A b}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {-b (A-2 C) \sec ^2(c+d x)+2 a C \sec (c+d x)+A b}{\sqrt {a+b \sec (c+d x)}}dx+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {-b (A-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a C \csc \left (c+d x+\frac {\pi }{2}\right )+A b}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 4546 |
\(\displaystyle \frac {1}{2} \left (\int \frac {A b+(b (A-2 C)+2 a C) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b (A-2 C) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\int \frac {A b+(b (A-2 C)+2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (A-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 4409 |
\(\displaystyle \frac {1}{2} \left (-b (A-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(2 C (a-b)+A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+A b \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left ((2 C (a-b)+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (A-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+A b \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 4271 |
\(\displaystyle \frac {1}{2} \left ((2 C (a-b)+A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (A-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {1}{2} \left (-b (A-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} (2 C (a-b)+A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \sqrt {a+b} (2 C (a-b)+A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} (A-2 C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}\right )+\frac {A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\) |
Input:
Int[Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2),x]
Output:
((2*(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) + (2*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) - (2*A*b*Sqrt[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*d))/2 + (A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*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)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(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)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[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]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(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 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
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] + Simp[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]
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*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[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] && Gt Q[m, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(898\) vs. \(2(323)=646\).
Time = 13.02 (sec) , antiderivative size = 899, normalized size of antiderivative = 2.55
Input:
int(cos(d*x+c)*(a+b*sec(d*x+c))^(1/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNV ERBOSE)
Output:
1/d*((-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x +c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b*EllipticPi(-csc(d*x+c)+c ot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1/(a+b )*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2) *a*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2* cos(d*x+c)-1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c )/(cos(d*x+c)+1))^(1/2)*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^( 1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x +c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*EllipticE(-csc(d*x+c)+co t(d*x+c),((a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*C*(1/(a+b)*( b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b* EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^2+4*co s(d*x+c)+2)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/ (cos(d*x+c)+1))^(1/2)*b*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/ 2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+ c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*EllipticF(-csc(d*x+c)+cot (d*x+c),((a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*C*(1/(a+b)*( b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b* EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+A*a*cos(d*x+c)^2*sin (d*x+c)+A*b*cos(d*x+c)*sin(d*x+c)+2*C*sin(d*x+c)*cos(d*x+c)*a+2*C*b*sin...
Timed out. \[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(1/2)*(A+C*sec(d*x+c)^2),x, algorith m="fricas")
Output:
Timed out
\[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \cos {\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))**(1/2)*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sqrt(a + b*sec(c + d*x))*cos(c + d*x), x)
\[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(1/2)*(A+C*sec(d*x+c)^2),x, algorith m="maxima")
Output:
integrate((C*sec(d*x + c)^2 + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c), x)
\[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(1/2)*(A+C*sec(d*x+c)^2),x, algorith m="giac")
Output:
integrate((C*sec(d*x + c)^2 + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c), x)
Timed out. \[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int(cos(c + d*x)*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/2),x)
Output:
int(cos(c + d*x)*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/2), x)
\[ \int \cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )d x \right ) a \] Input:
int(cos(d*x+c)*(a+b*sec(d*x+c))^(1/2)*(A+C*sec(d*x+c)^2),x)
Output:
int(sqrt(sec(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**2,x)*c + int(sqrt( sec(c + d*x)*b + a)*cos(c + d*x),x)*a