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

Optimal. Leaf size=283 \[ \frac{a^2 (80 A+90 B+67 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{240 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (176 A+150 B+133 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (176 A+150 B+133 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{128 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (176 A+150 B+133 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{128 d}+\frac{a (10 B+3 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{40 d}+\frac{C \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(a^(3/2)*(176*A + 150*B + 133*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(128*d) + (a^2*(176
*A + 150*B + 133*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(128*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(176*A + 150*B +
133*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(80*A + 90*B + 67*C)*Sec[c + d
*x]^(7/2)*Sin[c + d*x])/(240*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(10*B + 3*C)*Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c
+ d*x]]*Sin[c + d*x])/(40*d) + (C*Sec[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.748112, antiderivative size = 283, normalized size of antiderivative = 1., 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} \[ \frac{a^2 (80 A+90 B+67 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{240 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (176 A+150 B+133 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{192 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (176 A+150 B+133 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{128 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (176 A+150 B+133 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{128 d}+\frac{a (10 B+3 C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{40 d}+\frac{C \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/2)*(176*A + 150*B + 133*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(128*d) + (a^2*(176
*A + 150*B + 133*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(128*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(176*A + 150*B +
133*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(80*A + 90*B + 67*C)*Sec[c + d
*x]^(7/2)*Sin[c + d*x])/(240*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(10*B + 3*C)*Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c
+ d*x]]*Sin[c + d*x])/(40*d) + (C*Sec[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

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]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{\int \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{5}{2} a (2 A+C)+\frac{1}{2} a (10 B+3 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac{a (10 B+3 C) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{\int \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{5}{4} a^2 (16 A+10 B+11 C)+\frac{1}{4} a^2 (80 A+90 B+67 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac{a^2 (80 A+90 B+67 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a (10 B+3 C) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{1}{96} (a (176 A+150 B+133 C)) \int \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (80 A+90 B+67 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a (10 B+3 C) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{1}{128} (a (176 A+150 B+133 C)) \int \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (80 A+90 B+67 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a (10 B+3 C) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{1}{256} (a (176 A+150 B+133 C)) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (80 A+90 B+67 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a (10 B+3 C) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{(a (176 A+150 B+133 C)) \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^{3/2} (176 A+150 B+133 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{128 d}+\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (176 A+150 B+133 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (80 A+90 B+67 C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{240 d \sqrt{a+a \sec (c+d x)}}+\frac{a (10 B+3 C) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac{C \sec ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 3.89925, size = 211, normalized size = 0.75 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) (12 (880 A+1070 B+1273 C) \cos (c+d x)+4 (3280 A+3450 B+3059 C) \cos (2 (c+d x))+3520 A \cos (3 (c+d x))+2640 A \cos (4 (c+d x))+10480 A+3000 B \cos (3 (c+d x))+2250 B \cos (4 (c+d x))+11550 B+2660 C \cos (3 (c+d x))+1995 C \cos (4 (c+d x))+13313 C)+240 \sqrt{2} (176 A+150 B+133 C) \cos ^5(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{61440 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[(c + d*x)/2]*Sec[c + d*x]^(9/2)*Sqrt[a*(1 + Sec[c + d*x])]*(240*Sqrt[2]*(176*A + 150*B + 133*C)*ArcTanh
[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^5 + 4*(10480*A + 11550*B + 13313*C + 12*(880*A + 1070*B + 1273*C)*Cos[
c + d*x] + 4*(3280*A + 3450*B + 3059*C)*Cos[2*(c + d*x)] + 3520*A*Cos[3*(c + d*x)] + 3000*B*Cos[3*(c + d*x)] +
2660*C*Cos[3*(c + d*x)] + 2640*A*Cos[4*(c + d*x)] + 2250*B*Cos[4*(c + d*x)] + 1995*C*Cos[4*(c + d*x)])*Sin[(c
+ d*x)/2]))/(61440*d)

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Maple [B]  time = 0.417, size = 732, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7680/d*a*(2640*A*cos(d*x+c)^5*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2
)-2640*A*cos(d*x+c)^5*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*2^(1/2)+2250*B*c
os(d*x+c)^5*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2)-2250*B*cos(d*x+c)^
5*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*2^(1/2)+1995*C*cos(d*x+c)^5*arctan(1
/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2)-1995*C*cos(d*x+c)^5*arctan(1/4*2^(1/2)
*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*2^(1/2)+5280*A*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)*cos(
d*x+c)^4+4500*B*cos(d*x+c)^4*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+3990*C*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)*
cos(d*x+c)^4+3520*A*sin(d*x+c)*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))^(1/2)+3000*B*sin(d*x+c)*cos(d*x+c)^3*(-2/(cos(
d*x+c)+1))^(1/2)+2660*C*sin(d*x+c)*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))^(1/2)+1280*A*cos(d*x+c)^2*sin(d*x+c)*(-2/(
cos(d*x+c)+1))^(1/2)+2400*B*cos(d*x+c)^2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+2128*C*sin(d*x+c)*cos(d*x+c)^2*(
-2/(cos(d*x+c)+1))^(1/2)+960*B*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+1824*C*sin(d*x+c)*cos(d*x+c)*(-
2/(cos(d*x+c)+1))^(1/2)+768*C*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(1/cos
(d*x+c))^(5/2)*(-2/(cos(d*x+c)+1))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^2*(cos(d*x+c)^2-1)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 2.41543, size = 1543, normalized size = 5.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/7680*(15*((176*A + 150*B + 133*C)*a*cos(d*x + c)^5 + (176*A + 150*B + 133*C)*a*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*(176*A + 150*
B + 133*C)*a*cos(d*x + c)^4 + 10*(176*A + 150*B + 133*C)*a*cos(d*x + c)^3 + 8*(80*A + 150*B + 133*C)*a*cos(d*x
+ c)^2 + 48*(10*B + 19*C)*a*cos(d*x + c) + 384*C*a)*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*((176*A + 150*B + 133*C)*a*cos(d*x + c)^5 +
(176*A + 150*B + 133*C)*a*cos(d*x + c)^4)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s
qrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(15*(176*A + 150*B + 133*C)*a*co
s(d*x + c)^4 + 10*(176*A + 150*B + 133*C)*a*cos(d*x + c)^3 + 8*(80*A + 150*B + 133*C)*a*cos(d*x + c)^2 + 48*(1
0*B + 19*C)*a*cos(d*x + c) + 384*C*a)*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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{3}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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