\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [591]

Optimal result

Integrand size = 45, antiderivative size = 172 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 (12 A+20 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2*a^(3/2)*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/5*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec
(d*x+c)^(3/2)+2/15*a^2*(12*A+20*B+15*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a*(3*A+5*B)*
sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4171, 4102, 4100, 3886, 221} \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 (12 A+20 B+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (3 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(2*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(12*A + 20*B + 15*C)*Sqrt[Se
c[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 5*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c +
d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3886

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]

Rule 4100

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

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]

Rule 4171

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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (3 A+5 B)+\frac {5}{2} a C \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (12 A+20 B+15 C)+\frac {15}{4} a^2 C \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a^2 (12 A+20 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+(a C) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^2 (12 A+20 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(2 a C) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {2 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 (12 A+20 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.73 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^2 \left ((39 A+50 B+30 C+2 (9 A+5 B) \cos (c+d x)+3 A \cos (2 (c+d x))) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sin (c+d x)+30 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)\right )}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(a^2*((39*A + 50*B + 30*C + 2*(9*A + 5*B)*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[-((-1 + Sec[c + d*x])*Sec[
c + d*x])]*Sin[c + d*x] + 30*C*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x]))/(15*d*Sqrt[1 - Sec[c + d*x]]*Sqrt
[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.63

method	result	size
parts	\(\frac {2 A a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )+3 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 B a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )+5 \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}-\frac {C a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-\sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}\)	\(280\)
default	\(\frac {a \left (6 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+15 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+15 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+18 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+10 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+15 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+15 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+36 A \sin \left (d x +c \right )+50 B \sin \left (d x +c \right )+30 C \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\)	\(366\)

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/5*A/d*a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(5/2)*(sin(d*x+c)+3*tan(d*x+c)+6*sec(d*x+c)*tan(d
*x+c))+2/3*B/d*a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(sin(d*x+c)+5*tan(d*x+c))-C/d*a*(a*(
1+sec(d*x+c)))^(1/2)/sec(d*x+c)^(1/2)*(-arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1
))^(1/2))*(-1/(cos(d*x+c)+1))^(1/2)-(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)
+1)/(-1/(cos(d*x+c)+1))^(1/2))+2*cot(d*x+c)-2*csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.38 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {15 \, {\left (C a \cos \left (d x + c\right ) + C a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (3 \, A a \cos \left (d x + c\right )^{3} + {\left (9 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (18 \, A + 25 \, B + 15 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {15 \, {\left (C a \cos \left (d x + c\right ) + C a\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (3 \, A a \cos \left (d x + c\right )^{3} + {\left (9 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (18 \, A + 25 \, B + 15 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/30*(15*(C*a*cos(d*x + c) + C*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*
cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x
+ c)^3 + cos(d*x + c)^2)) + 4*(3*A*a*cos(d*x + c)^3 + (9*A + 5*B)*a*cos(d*x + c)^2 + (18*A + 25*B + 15*C)*a*co
s(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/1
5*(15*(C*a*cos(d*x + c) + C*a)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x
+ c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(3*A*a*cos(d*x + c)^3 + (9*A + 5*B)*a*cos(d
*x + c)^2 + (18*A + 25*B + 15*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos
(d*x + c)))/(d*cos(d*x + c) + d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 522 vs. \(2 (148) = 296\).

Time = 0.52 (sec) , antiderivative size = 522, normalized size of antiderivative = 3.03 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \, \sqrt {2} {\left (20 \, a \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 20 \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 5 \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 2 \, a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 20 \, a \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 15 \, \sqrt {2} {\left (\sqrt {2} a \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \sqrt {2} a \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) + \sqrt {2} a \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \sqrt {2} a \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) + 8 \, a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a} + 20 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{60 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/60*(3*sqrt(2)*(20*a*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*a*
cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 20*a*cos(5/2*d*x + 5/2*c)*
sin(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*a*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/
2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 2*a*sin(5/2*d*x + 5/2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c),
cos(5/2*d*x + 5/2*c))) + 20*a*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 15*sqr
t(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*s
qrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(
2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin
(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*c
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/
2*c) + 2) + 8*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a) + 20*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x
+ 1/2*c))*B*sqrt(a))/d

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2),x)

[Out]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2), x)