\(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [189]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 158 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/3*A*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/3*a^2*(A-3*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+4*a^2*A
*(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*(2*A+3*B)*(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 = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4102, 4082, 3872, 3856, 2719, 2720} \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 a^2 (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d \sqrt {\sec (c+d x)}}+\frac {4 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d} \]

[In]

Int[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(3/2),x]

[Out]

(4*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^2*(2*A + 3*B)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*a^2*(A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(3*d) + (2*A*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

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 4102

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

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

Mathematica [C] (verified)

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

Time = 3.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (-i \cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (12 A-\frac {24 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+\frac {8 (2 A+3 B) e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+2 i A \sin (c+d x)+6 i B \tan (c+d x)\right )}{3 d \sqrt {\sec (c+d x)}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(3/2),x]

[Out]

(a^2*(Cos[c/2] - I*Sin[c/2])*((-I)*Cos[c/2] + Sin[c/2])*(12*A - (24*A*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2
*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (8*(2*A + 3*B)*E^(I*(c + d*x))*Hypergeometric2F1[1/4, 1/2, 5/
4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (2*I)*A*Sin[c + d*x] + (6*I)*B*Tan[c + d*x]))/(3*d*S
qrt[Sec[c + d*x]])

Maple [A] (verified)

Time = 21.19 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.55

method result size
default \(-\frac {4 a^{2} \left (2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(245\)
parts \(-\frac {2 \left (A \,a^{2}+2 B \,a^{2}\right ) \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}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}+\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \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}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 A \,a^{2} \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 (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 B \,a^{2} \left (-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}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\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}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(657\)

[In]

int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-4/3*a^2*(2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-A*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+2*A*(2*sin(1/2
*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^2+3*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(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 = 166, normalized size of antiderivative = 1.05 \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (i \, \sqrt {2} {\left (2 \, A + 3 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (2 \, A + 3 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} A a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} A a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (A a^{2} \cos \left (d x + c\right ) + 3 \, B a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-2/3*(I*sqrt(2)*(2*A + 3*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - I*sqrt(2)*(2*A + 3
*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*A*a^2*weierstrassZeta(-4, 0, w
eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*A*a^2*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (A*a^2*cos(d*x + c) + 3*B*a^2)*sin(d*x + c)/sqrt(cos(d*x +
 c)))/d

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=a^{2} \left (\int \frac {A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int A \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \frac {B}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 2 B \sqrt {\sec {\left (c + d x \right )}}\, dx + \int B \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))**2*(A+B*sec(d*x+c))/sec(d*x+c)**(3/2),x)

[Out]

a**2*(Integral(A/sec(c + d*x)**(3/2), x) + Integral(2*A/sqrt(sec(c + d*x)), x) + Integral(A*sqrt(sec(c + d*x))
, x) + Integral(B/sqrt(sec(c + d*x)), x) + Integral(2*B*sqrt(sec(c + d*x)), x) + Integral(B*sec(c + d*x)**(3/2
), x))

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(3/2), x)

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sec(d*x + c)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(3/2), x)