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

Optimal. Leaf size=191 \[ \frac{2 (13 A-5 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-5 B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/
(Sqrt[a]*d)) + (2*A*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - (2*(A - 5*B)*Sin[c + d*x
])/(15*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*(13*A - 5*B + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x]
)/(15*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.570631, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4086, 4022, 4013, 3808, 206} \[ \frac{2 (13 A-5 B+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-5 B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/
(Sqrt[a]*d)) + (2*A*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - (2*(A - 5*B)*Sin[c + d*x
])/(15*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (2*(13*A - 5*B + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x]
)/(15*d*Sqrt[a + a*Sec[c + d*x]])

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 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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 - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 3808

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{-\frac{1}{2} a (A-5 B)+\frac{1}{2} a (4 A+5 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} a^2 (13 A-5 B+15 C)-\frac{1}{2} a^2 (A-5 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 (13 A-5 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+(-A+B-C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 (13 A-5 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{(2 (A-B+C)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 (13 A-5 B+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.885045, size = 155, normalized size = 0.81 \[ -\frac{4 \cos \left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (15 (A-B+C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+20 (A+B) \sin ^3\left (\frac{1}{2} (c+d x)\right )-30 (A+C) \sin \left (\frac{1}{2} (c+d x)\right )-24 A \sin ^5\left (\frac{1}{2} (c+d x)\right )\right )}{15 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(15*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] - 30*(A
+ C)*Sin[(c + d*x)/2] + 20*(A + B)*Sin[(c + d*x)/2]^3 - 24*A*Sin[(c + d*x)/2]^5))/(15*d*(A + 2*C + 2*B*Cos[c
+ d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2)*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.427, size = 263, normalized size = 1.4 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,ad\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}A\sin \left ( dx+c \right ) -6\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-15\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}B\sin \left ( dx+c \right ) +15\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sin \left ( dx+c \right ) +8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-10\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-28\,A\cos \left ( dx+c \right ) +20\,B\cos \left ( dx+c \right ) -30\,C\cos \left ( dx+c \right ) +26\,A-10\,B+30\,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+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/15/d/a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(15*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x
+c)+1))^(1/2)*A*sin(d*x+c)-6*A*cos(d*x+c)^3-15*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c
)+1))^(1/2)*B*sin(d*x+c)+15*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d
*x+c)+8*A*cos(d*x+c)^2-10*B*cos(d*x+c)^2-28*A*cos(d*x+c)+20*B*cos(d*x+c)-30*C*cos(d*x+c)+26*A-10*B+30*C)*cos(d
*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)

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Maxima [B]  time = 2.45713, size = 1002, normalized size = 5.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/60*(sqrt(2)*(60*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*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) - 60*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*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))) - 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5
*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x
+ 5/2*c))) + 1) + 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arctan2(sin(
5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 - 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) +
1) + 6*sin(5/2*d*x + 5/2*c) - 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 60*sin(1/5*arct
an2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A/sqrt(a) - 10*(3*sqrt(2)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*
c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*sqrt(2)*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2)*log(cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2
+ sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co
s(3/2*d*x + 3/2*c))) + 1) + 3*sqrt(2)*log(cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin
(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*
d*x + 3/2*c))) + 1) - 2*sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2
*d*x + 3/2*c))))*B/sqrt(a) - 30*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x +
1/2*c) + 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 4*s
qrt(2)*sin(1/2*d*x + 1/2*c))*C/sqrt(a))/d

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Fricas [A]  time = 0.544631, size = 1069, normalized size = 5.6 \begin{align*} \left [\frac{\frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} + \frac{2 \, \sqrt{2} \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 )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}} + \frac{4 \,{\left (3 \, A \cos \left (d x + c\right )^{3} -{\left (A - 5 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (13 \, A - 5 \, B + 15 \, C\right )} \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 (a d \cos \left (d x + c\right ) + a d\right )}}, \frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\right )} \sqrt{-\frac{1}{a}} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + \frac{2 \,{\left (3 \, A \cos \left (d x + c\right )^{3} -{\left (A - 5 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (13 \, A - 5 \, B + 15 \, C\right )} \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 (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*co
s(d*x + c) + 1))/sqrt(a) + 4*(3*A*cos(d*x + c)^3 - (A - 5*B)*cos(d*x + c)^2 + (13*A - 5*B + 15*C)*cos(d*x + c)
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d), 1/15*(15*
sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) + 2*(3*A*cos(d*x + c)^3 - (A - 5*B)*cos(d*x + c)^2 + (13*
A - 5*B + 15*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*co
s(d*x + c) + a*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+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\sqrt{a \sec \left (d x + c\right ) + a} \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+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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