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

Optimal. Leaf size=223 \[ \frac {a^{5/2} (2 B+5 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (64 A+70 B+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (16 A+10 B-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a (A+B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

a^(5/2)*(2*B+5*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))^(5/2)*sin(d*x+c)
/d/sec(d*x+c)^(3/2)+2/3*a*(A+B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+1/15*a^3*(64*A+70*B+15*C)
*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)-1/15*a^2*(16*A+10*B-15*C)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a
+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.72, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4086, 4017, 4018, 4015, 3801, 215} \[ \frac {a^3 (64 A+70 B+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (16 A+10 B-15 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {a^{5/2} (2 B+5 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a (A+B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(2*B + 5*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^3*(64*A + 70*B + 15*C)*S
qrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(16*A + 10*B - 15*C)*Sqrt[Sec[c + d*x]]
*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*(A + B)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sq
rt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

Rule 215

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 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), 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 4015

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 4017

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 4018

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] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 4086

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*} \int \frac {(a+a \sec (c+d x))^{5/2} \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 (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {5}{2} a (A+B)-\frac {1}{2} a (2 A-5 C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac {2 a (A+B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (8 A+10 B+5 C)-\frac {1}{4} a^2 (16 A+10 B-15 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a}\\ &=-\frac {a^2 (16 A+10 B-15 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (A+B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{5/2} \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}{8} a^3 (64 A+70 B+15 C)+\frac {15}{8} a^3 (2 B+5 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{15 a}\\ &=\frac {a^3 (64 A+70 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (16 A+10 B-15 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (A+B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{2} \left (a^2 (2 B+5 C)\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (64 A+70 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (16 A+10 B-15 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (A+B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\left (a^2 (2 B+5 C)\right ) \operatorname {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 {a^{5/2} (2 B+5 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^3 (64 A+70 B+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (16 A+10 B-15 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a (A+B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 149, normalized size = 0.67 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \sqrt {a (\sec (c+d x)+1)} \left (2 \sin \left (\frac {1}{2} (c+d x)\right ) ((181 A+160 B+60 C) \cos (c+d x)+2 (14 A+5 B) \cos (2 (c+d x))+3 A \cos (3 (c+d x))+28 A+10 B+30 C)+30 \sqrt {2} (2 B+5 C) \cos (c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])]*(30*Sqrt[2]*(2*B + 5*C)*ArcTanh[Sqrt[2]*Si
n[(c + d*x)/2]]*Cos[c + d*x] + 2*(28*A + 10*B + 30*C + (181*A + 160*B + 60*C)*Cos[c + d*x] + 2*(14*A + 5*B)*Co
s[2*(c + d*x)] + 3*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(60*d)

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fricas [A]  time = 0.66, size = 470, normalized size = 2.11 \[ \left [\frac {15 \, {\left ({\left (2 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (2 \, B + 5 \, C\right )} a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (43 \, A + 40 \, B + 15 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {15 \, {\left ({\left (2 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (2 \, B + 5 \, C\right )} a^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (43 \, A + 40 \, B + 15 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(15*((2*B + 5*C)*a^2*cos(d*x + c) + (2*B + 5*C)*a^2)*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*(6*A*a^2*cos(d*x + c)^3 + 2*(14*A + 5*B)*a^2*cos(d*x + c
)^2 + 2*(43*A + 40*B + 15*C)*a^2*cos(d*x + c) + 15*C*a^2)*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), 1/30*(15*((2*B + 5*C)*a^2*cos(d*x + c) + (2*B + 5*C)*a^2)*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*(6*A*a^2*cos(d*x + c)^3 + 2*(14*A + 5*B)*a^2*cos(d*x + c)^2 + 2*(43*A + 40*B + 15*C
)*a^2*cos(d*x + c) + 15*C*a^2)*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)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.68, size = 420, normalized size = 1.88 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (30 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-30 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )+75 C \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-75 C \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+24 A \left (\cos ^{4}\left (d x +c \right )\right )+88 A \left (\cos ^{3}\left (d x +c \right )\right )+40 B \left (\cos ^{3}\left (d x +c \right )\right )+232 A \left (\cos ^{2}\left (d x +c \right )\right )+280 B \left (\cos ^{2}\left (d x +c \right )\right )+120 C \left (\cos ^{2}\left (d x +c \right )\right )-344 A \cos \left (d x +c \right )-320 B \cos \left (d x +c \right )-60 C \cos \left (d x +c \right )-60 C \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} a^{2}}{60 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(30*B*sin(d*x+c)*cos(d*x+c)*2^(1/2)*(-2/(1+cos(d*x+c)))^(1/2)*arct
an(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))-30*B*sin(d*x+c)*cos(d*x+c)*2^(1/2)*(-2/(1+
cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))+75*C*2^(1/2)*sin(d*
x+c)*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/
2))-75*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))
*(-2/(1+cos(d*x+c)))^(1/2)+24*A*cos(d*x+c)^4+88*A*cos(d*x+c)^3+40*B*cos(d*x+c)^3+232*A*cos(d*x+c)^2+280*B*cos(
d*x+c)^2+120*C*cos(d*x+c)^2-344*A*cos(d*x+c)-320*B*cos(d*x+c)-60*C*cos(d*x+c)-60*C)*cos(d*x+c)^2*(1/cos(d*x+c)
)^(5/2)/sin(d*x+c)*a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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