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

[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))

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Rubi [A]  time = 0.71922, antiderivative size = 223, normalized size of antiderivative = 1., 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 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])

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 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 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 \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*}

Mathematica [A]  time = 1.1362, 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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Maple [B]  time = 0.36, size = 420, normalized size = 1.9 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{60\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( -30\,B\cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) +30\,B\cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) -75\,C\cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) +75\,C\cos \left ( dx+c \right ) \sqrt{2}\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) +24\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+88\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+40\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+232\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+280\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+120\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-344\,A\cos \left ( dx+c \right ) -320\,B\cos \left ( dx+c \right ) -60\,C\cos \left ( dx+c \right ) -60\,C \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}} \end{align*}

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^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-30*B*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)
*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+30*B*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-
2/(cos(d*x+c)+1))^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-75*C*cos(d*x+c
)*2^(1/2)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(
d*x+c)))+75*C*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(
1/2)*(cos(d*x+c)+1+sin(d*x+c)))+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)

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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((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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Fricas [A]  time = 0.819469, size = 1214, normalized size = 5.44 \begin{align*} \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 ] \end{align*}

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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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((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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)