Integrand size = 33, antiderivative size = 153 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {(A-3 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {(A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:
(A-3*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^ (1/2)/a/d+(A-B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*se c(d*x+c)^(1/2)/a/d-(A-3*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d+(A-B)*sec(d*x+c )^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.50 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.75 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) (A+B \sec (c+d x)) \left (-2 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+6 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+12 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-12 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-6 \sqrt {\sec (c+d x)} \left (2 (A-3 B) \cos (d x) \csc (c)+2 (-A+B) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{6 a d (B+A \cos (c+d x)) (1+\sec (c+d x))} \] Input:
Integrate[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x ]
Output:
(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x])*((-2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x ))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqr t[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometr ic2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (6*Sqrt[2]*B*Sqrt [E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]* Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c) )*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 12* A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 12*B*S qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 6*Sqrt[Se c[c + d*x]]*(2*(A - 3*B)*Cos[d*x]*Csc[c] + 2*(-A + B)*Tan[(c + d*x)/2])))/ (6*a*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x]))
Time = 0.80 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4507, 27, 3042, 4274, 3042, 4255, 3042, 4258, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a \sec (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} (a (A-B)-a (A-3 B) \sec (c+d x))dx}{a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {\sec (c+d x)} (a (A-B)-a (A-3 B) \sec (c+d x))dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (A-B)-a (A-3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (A-B) \int \sqrt {\sec (c+d x)}dx-a (A-3 B) \int \sec ^{\frac {3}{2}}(c+d x)dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (A-B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-a (A-3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {a (A-B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-a (A-3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (A-B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-a (A-3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-a (A-3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (A-3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (A-3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {2 a (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-a (A-3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
Input:
Int[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
Output:
((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((2*a *(A - B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/ d - a*(A - 3*B)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec [c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/(2*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(317\) vs. \(2(144)=288\).
Time = 2.04 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.08
| method | result | size |
| default | \(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A -3 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A -5 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(318\) |
Input:
int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*(-cos(1/2*d*x +1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+ 1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*EllipticF(cos(1/2*d*x+ 1/2*c),2^(1/2))-A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))+3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(-2*sin (1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-3*B)*sin(1/2*d*x+1/2*c)^4 -(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-5*B)*sin(1/2*d*x+ 1/2*c)^2)/cos(1/2*d*x+1/2*c)/sin(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^2- 1)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 3 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (-i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 3 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left ({\left (A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, B\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm= "fricas")
Output:
1/2*((sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstras sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - I*B)*cos (d*x + c) + sqrt(2)*(I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (sqrt(2)*(I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(I*A - 3* I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s in(d*x + c))) + (sqrt(2)*(-I*A + 3*I*B)*cos(d*x + c) + sqrt(2)*(-I*A + 3*I *B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c))) - 2*((A - 3*B)*cos(d*x + c) - 2*B)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{\frac {5}{2}}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)**(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
Output:
(Integral(A*sec(c + d*x)**(3/2)/(sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**(5/2)/(sec(c + d*x) + 1), x))/a
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \] Input:
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \] Input:
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x)),x)
Output:
int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x)), x)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) a}{a} \] Input:
int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
Output:
(int((sqrt(sec(c + d*x))*sec(c + d*x)**2)/(sec(c + d*x) + 1),x)*b + int((s qrt(sec(c + d*x))*sec(c + d*x))/(sec(c + d*x) + 1),x)*a)/a