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

Optimal. Leaf size=281 \[ \frac {a^{5/2} (400 A+326 B+283 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a^3 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{960 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (400 A+326 B+283 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{128 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{240 d}+\frac {a (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{8 d}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d} \]

[Out]

1/128*a^(5/2)*(400*A+326*B+283*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/8*a*(2*B+C)*sec(d*x+c
)^(5/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/5*C*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/128*a
^3*(400*A+326*B+283*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/960*a^3*(1040*A+950*B+787*C)*sec
(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/240*a^2*(80*A+110*B+79*C)*sec(d*x+c)^(5/2)*sin(d*x+c)*(a+a
*sec(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.86, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4088, 4018, 4016, 3803, 3801, 215} \[ \frac {a^3 (1040 A+950 B+787 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{960 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (400 A+326 B+283 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{128 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (80 A+110 B+79 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{240 d}+\frac {a^{5/2} (400 A+326 B+283 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a (2 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{8 d}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(400*A + 326*B + 283*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(128*d) + (a^3*(400
*A + 326*B + 283*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(128*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(1040*A + 950*B +
787*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(960*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(80*A + 110*B + 79*C)*Sec[c +
d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(240*d) + (a*(2*B + C)*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*
x])^(3/2)*Sin[c + d*x])/(8*d) + (C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d)

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 3803

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

Rule 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, 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 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 4088

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*} \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (10 A+3 C)+\frac {5}{2} a (2 B+C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac {a (2 B+C) \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (80 A+30 B+39 C)+\frac {1}{4} a^2 (80 A+110 B+79 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac {a^2 (80 A+110 B+79 C) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (2 B+C) \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {\int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{8} a^3 (240 A+170 B+157 C)+\frac {1}{8} a^3 (1040 A+950 B+787 C) \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac {a^3 (1040 A+950 B+787 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (80 A+110 B+79 C) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (2 B+C) \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{128} \left (a^2 (400 A+326 B+283 C)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (400 A+326 B+283 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1040 A+950 B+787 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (80 A+110 B+79 C) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (2 B+C) \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{256} \left (a^2 (400 A+326 B+283 C)\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (400 A+326 B+283 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1040 A+950 B+787 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (80 A+110 B+79 C) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (2 B+C) \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac {\left (a^2 (400 A+326 B+283 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 )}{128 d}\\ &=\frac {a^{5/2} (400 A+326 B+283 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (400 A+326 B+283 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1040 A+950 B+787 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (80 A+110 B+79 C) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (2 B+C) \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {C \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.49, size = 213, normalized size = 0.76 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (\sec (c+d x)+1)} \left (4 \sin \left (\frac {1}{2} (c+d x)\right ) (12 (1360 A+1950 B+2343 C) \cos (c+d x)+4 (6640 A+6730 B+6509 C) \cos (2 (c+d x))+5440 A \cos (3 (c+d x))+6000 A \cos (4 (c+d x))+20560 A+6520 B \cos (3 (c+d x))+4890 B \cos (4 (c+d x))+22030 B+5660 C \cos (3 (c+d x))+4245 C \cos (4 (c+d x))+24863 C)+240 \sqrt {2} (400 A+326 B+283 C) \cos ^5(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{61440 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^(9/2)*Sqrt[a*(1 + Sec[c + d*x])]*(240*Sqrt[2]*(400*A + 326*B + 283*C)*ArcTa
nh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^5 + 4*(20560*A + 22030*B + 24863*C + 12*(1360*A + 1950*B + 2343*C)*C
os[c + d*x] + 4*(6640*A + 6730*B + 6509*C)*Cos[2*(c + d*x)] + 5440*A*Cos[3*(c + d*x)] + 6520*B*Cos[3*(c + d*x)
] + 5660*C*Cos[3*(c + d*x)] + 6000*A*Cos[4*(c + d*x)] + 4890*B*Cos[4*(c + d*x)] + 4245*C*Cos[4*(c + d*x)])*Sin
[(c + d*x)/2]))/(61440*d)

________________________________________________________________________________________

fricas [A]  time = 1.16, size = 588, normalized size = 2.09 \[ \left [\frac {15 \, {\left ({\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{4}\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 (15 \, {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (272 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (80 \, A + 230 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (10 \, B + 29 \, C\right )} a^{2} \cos \left (d x + c\right ) + 384 \, 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 )}}}{7680 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}, \frac {15 \, {\left ({\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{4}\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 (15 \, {\left (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (272 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (80 \, A + 230 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (10 \, B + 29 \, C\right )} a^{2} \cos \left (d x + c\right ) + 384 \, 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 )}}}{3840 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/7680*(15*((400*A + 326*B + 283*C)*a^2*cos(d*x + c)^5 + (400*A + 326*B + 283*C)*a^2*cos(d*x + c)^4)*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*(15*(400*A +
326*B + 283*C)*a^2*cos(d*x + c)^4 + 10*(272*A + 326*B + 283*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a
^2*cos(d*x + c)^2 + 48*(10*B + 29*C)*a^2*cos(d*x + c) + 384*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)^5 + d*cos(d*x + c)^4), 1/3840*(15*((400*A + 326*B + 283*C)*a^2*c
os(d*x + c)^5 + (400*A + 326*B + 283*C)*a^2*cos(d*x + c)^4)*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*(15*(400*A + 3
26*B + 283*C)*a^2*cos(d*x + c)^4 + 10*(272*A + 326*B + 283*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a^
2*cos(d*x + c)^2 + 48*(10*B + 29*C)*a^2*cos(d*x + c) + 384*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)^5 + d*cos(d*x + c)^4)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maple [B]  time = 2.19, size = 734, normalized size = 2.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7680/d*(6000*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)
-6000*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)+4890*B*ar
ctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)-4890*B*arctan(1/4*(
-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)+4245*C*2^(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))*cos(d*x+c)^5-4245*C*2^(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))*cos(d*x+c)^5-12000*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*si
n(d*x+c)-9780*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin(d*x+c)-8490*C*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^
4*sin(d*x+c)-5440*A*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-6520*B*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+co
s(d*x+c)))^(1/2)-5660*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^3-1280*A*sin(d*x+c)*cos(d*x+c)^2*(-2/(
1+cos(d*x+c)))^(1/2)-3680*B*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)-4528*C*(-2/(1+cos(d*x+c)))^(1/2)
*sin(d*x+c)*cos(d*x+c)^2-960*B*sin(d*x+c)*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-2784*C*(-2/(1+cos(d*x+c)))^(1/2
)*sin(d*x+c)*cos(d*x+c)-768*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1/cos
(d*x+c))^(3/2)*(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2/cos(d*x+c)^3*(cos(d*x+c)^2-1)*a^2

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________