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

Optimal. Leaf size=188 \[ \frac{(A+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{3 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(A+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(-3*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + ((A + 9*C)*ArcTanh[(Sqrt[a]*Sqrt
[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((A + 3*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*d*Sq
rt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.562369, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {4085, 4021, 4023, 3808, 206, 3801, 215} \[ \frac{(A+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{3 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(A+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + ((A + 9*C)*ArcTanh[(Sqrt[a]*Sqrt
[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((A + 3*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*d*Sq
rt[a + a*Sec[c + d*x]])

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(
2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b*(
2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4021

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(f*(m + n
)), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m +
n) + a*B*m)*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] && GtQ[n, 1]

Rule 4023

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

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])

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{\sec ^{\frac{3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-\frac{1}{2} a (A-3 C)-a (A+3 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}-\frac{\int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{2} a^2 (A+3 C)+3 a^2 C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^3}\\ &=-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}-\frac{(3 C) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{2 a^2}+\frac{(A+9 C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{(3 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 )}{a^2 d}-\frac{(A+9 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 )}{2 a d}\\ &=-\frac{3 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac{(A+9 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 )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 7.0676, size = 800, normalized size = 4.26 \[ \frac{(\sec (c+d x)+1)^{3/2} \left (C \sec ^2(c+d x)+A\right ) \left (\frac{(A+3 C) \cos ^2(c+d x) \left (\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt{2} \sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1} \sqrt{\sec (c+d x)}+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt{2} \sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1} \sqrt{\sec (c+d x)}+1\right )\right ) (\sec (c+d x)+1) \sqrt{\sec ^2(c+d x)-1} \sin (c+d x)}{2 d (\cos (c+d x)+1) \sqrt{2-2 \cos ^2(c+d x)} \sqrt{1-\cos ^2(c+d x)}}-\frac{3 C \cos ^2(c+d x) \left (-8 \log (\sec (c+d x)+1)+8 \log \left (\sec ^{\frac{3}{2}}(c+d x)+\sqrt{\sec (c+d x)}+\sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1}\right )+\sqrt{2} \left (\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt{2} \sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1} \sqrt{\sec (c+d x)}+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt{2} \sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1} \sqrt{\sec (c+d x)}+1\right )\right )\right ) (\sec (c+d x)+1) \sqrt{\sec ^2(c+d x)-1} \sin (c+d x)}{2 d (\cos (c+d x)+1) \left (1-\cos ^2(c+d x)\right )}\right ) \cos ^2(c+d x)}{2 (\cos (2 c+2 d x) A+A+2 C) (a (\sec (c+d x)+1))^{3/2}}+\frac{\sqrt{(\cos (c+d x)+1) \sec (c+d x)} (\sec (c+d x)+1)^{3/2} \left (C \sec ^2(c+d x)+A\right ) \left (\frac{\sec \left (\frac{c}{2}\right ) \left (-A \sin \left (\frac{d x}{2}\right )-C \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\sec \left (\frac{c}{2}\right ) \left (-A \sin \left (\frac{c}{2}\right )-C \sin \left (\frac{c}{2}\right )\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+3 C \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{(A+3 C) \tan \left (\frac{c}{2}\right )}{d}\right )}{(\cos (2 c+2 d x) A+A+2 C) \sec ^{\frac{3}{2}}(c+d x) (a (\sec (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^2*(1 + Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*(((A + 3*C)*Cos[c + d*x]^2*(Log[1 - 2*Sec[c +
d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]^2]] - Log
[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c +
d*x]^2]])*(1 + Sec[c + d*x])*Sqrt[-1 + Sec[c + d*x]^2]*Sin[c + d*x])/(2*d*(1 + Cos[c + d*x])*Sqrt[2 - 2*Cos[c
+ d*x]^2]*Sqrt[1 - Cos[c + d*x]^2]) - (3*C*Cos[c + d*x]^2*(-8*Log[1 + Sec[c + d*x]] + 8*Log[Sqrt[Sec[c + d*x]
] + Sec[c + d*x]^(3/2) + Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]^2]] + Sqrt[2]*(-Log[1 - 2*Sec[c + d*x]
- 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]^2]] + Log[1 -
2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]
^2]]))*(1 + Sec[c + d*x])*Sqrt[-1 + Sec[c + d*x]^2]*Sin[c + d*x])/(2*d*(1 + Cos[c + d*x])*(1 - Cos[c + d*x]^2)
)))/(2*(A + 2*C + A*Cos[2*c + 2*d*x])*(a*(1 + Sec[c + d*x]))^(3/2)) + (Sqrt[(1 + Cos[c + d*x])*Sec[c + d*x]]*(
1 + Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*((Sec[c/2]*Sec[c/2 + (d*x)/2]^2*(-(A*Sin[c/2]) - C*Sin[c/2]))/(
2*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) - C*Sin[(d*x)/2]))/(2*d) + (Sec[c/2]*Sec[c/2 + (d*x)/
2]*(A*Sin[(d*x)/2] + 3*C*Sin[(d*x)/2]))/d + ((A + 3*C)*Tan[c/2])/d))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d
*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.388, size = 370, normalized size = 2. \begin{align*}{\frac{\cos \left ( dx+c \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }{4\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( -3\,C\sin \left ( dx+c \right ) \sqrt{2}\cos \left ( dx+c \right ) \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 ) +3\,C\sin \left ( dx+c \right ) \sqrt{2}\cos \left ( dx+c \right ) \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 ) +A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ({\frac{\sin \left ( dx+c \right ) }{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) -A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+9\,C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) -3\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+A\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+C\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+2\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/a^2*(-3*C*sin(d*x+c)*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d
*x+c)))+3*C*sin(d*x+c)*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c
)))+A*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-A*cos(d*x+c)^2*(-2/(cos(d*x+c)+1)
)^(1/2)+9*C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-3*C*cos(d*x+c)^2*(-2/(cos(d
*x+c)+1))^(1/2)+A*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+2*C*(-2/(cos(d*x
+c)+1))^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)/s
in(d*x+c)^3*(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)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.702915, size = 1658, normalized size = 8.82 \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)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(2)*((A + 9*C)*cos(d*x + c)^2 + 2*(A + 9*C)*cos(d*x + c) + A + 9*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 -
2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c)
- 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 6*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*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*((A + 3*C)*cos(d*x + c)
+ 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2
*d*cos(d*x + c) + a^2*d), -1/4*(sqrt(2)*((A + 9*C)*cos(d*x + c)^2 + 2*(A + 9*C)*cos(d*x + c) + A + 9*C)*sqrt(-
a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 6*(C
*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqr
t(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) - 2*((A + 3*C)*cos(d*x + c) + 2*C)*sqr
t((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x
+ c) + a^2*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(sec(d*x+c)**(3/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),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} + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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