\(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [543]

Optimal result

Integrand size = 43, antiderivative size = 214 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {4 a^2 (5 B+4 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (3 A+2 B+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (15 A+25 B+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]

[Out]

2/15*a^2*(15*A+25*B+17*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*C*(a+a*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d
+2/15*(5*B+4*C)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^2*(5*B+4*C)*(cos(1/2*d*x+1/2*c)^2)^(1
/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/3*a^2*(3*A+
2*B+C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*
sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4173, 4103, 4082, 3872, 3856, 2719, 2720} \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a^2 (15 A+25 B+17 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {4 a^2 (3 A+2 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 (5 B+4 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{15 d}-\frac {4 a^2 (5 B+4 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]

[In]

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

[Out]

(-4*a^2*(5*B + 4*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(3*A + 2*B
+ C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^2*(15*A + 25*B + 17*C)*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + (2
*(5*B + 4*C)*Sqrt[Sec[c + d*x]]*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(15*d)

Rule 2719

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]

Rule 2720

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]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e,
f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4103

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

Rule 4173

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{2} a (5 A-C)+\frac {1}{2} a (5 B+4 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{5 a} \\ & = \frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {4 \int \frac {(a+a \sec (c+d x)) \left (\frac {1}{4} a^2 (15 A-5 B-7 C)+\frac {1}{4} a^2 (15 A+25 B+17 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a^2 (15 A+25 B+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {8 \int \frac {-\frac {3}{4} a^3 (5 B+4 C)+\frac {5}{4} a^3 (3 A+2 B+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a^2 (15 A+25 B+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{3} \left (2 a^2 (3 A+2 B+C)\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (2 a^2 (5 B+4 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^2 (15 A+25 B+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{3} \left (2 a^2 (3 A+2 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (2 a^2 (5 B+4 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a^2 (5 B+4 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (3 A+2 B+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^2 (15 A+25 B+17 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 B+4 C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.47 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.24 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^2 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-90 i B \cos (c+d x)-72 i C \cos (c+d x)-30 i B \cos (3 (c+d x))-24 i C \cos (3 (c+d x))+40 (3 A+2 B+C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 i (5 B+4 C) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+15 A \sin (c+d x)+30 B \sin (c+d x)+36 C \sin (c+d x)+10 B \sin (2 (c+d x))+20 C \sin (2 (c+d x))+15 A \sin (3 (c+d x))+30 B \sin (3 (c+d x))+24 C \sin (3 (c+d x))\right )}{30 d} \]

[In]

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

[Out]

(a^2*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((-90*I)*B*Cos[c + d*x] - (72*I)*C*Cos[c + d*x] - (30*I)*B*Cos
[3*(c + d*x)] - (24*I)*C*Cos[3*(c + d*x)] + 40*(3*A + 2*B + C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] +
((2*I)*(5*B + 4*C)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(
I*(c + d*x)) + 15*A*Sin[c + d*x] + 30*B*Sin[c + d*x] + 36*C*Sin[c + d*x] + 10*B*Sin[2*(c + d*x)] + 20*C*Sin[2*
(c + d*x)] + 15*A*Sin[3*(c + d*x)] + 30*B*Sin[3*(c + d*x)] + 24*C*Sin[3*(c + d*x)]))/(30*d*E^(I*d*x))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(880\) vs. \(2(242)=484\).

Time = 5.37 (sec) , antiderivative size = 881, normalized size of antiderivative = 4.12

method	result	size
default	\(\text {Expression too large to display}\)	\(881\)
parts	\(\text {Expression too large to display}\)	\(1065\)

[In]

int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)+2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)
^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(c
os(1/2*d*x+1/2*c),2^(1/2)))+2/5*C/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1
/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^4+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/
2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)+8*(1/4*B+1/2*C)*(-1/6*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)/(cos(1/2*d*x+1
/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+8*(1/4*A+1/2*B+1/4*C)/sin(1/2*d*x+1/2*c)^2/(2*
sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))
)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, {\left (5 \, A + 10 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-2/15*(5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)
) - 5*I*sqrt(2)*(3*A + 2*B + C)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) +
3*I*sqrt(2)*(5*B + 4*C)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c))) - 3*I*sqrt(2)*(5*B + 4*C)*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c))) - (3*(5*A + 10*B + 8*C)*a^2*cos(d*x + c)^2 + 5*(B + 2*C)*a^2*cos(d*x + c) + 3
*C*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

int(((a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(1/2),x)

[Out]

int(((a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(1/2), x)