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

Optimal. Leaf size=169 \[ \frac{a^2 (8 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{3 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}-\frac{a (2 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}{3 d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt{\sec (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.467308, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4087, 4018, 4015, 3801, 215} \[ \frac{a^2 (8 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{3 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}-\frac{a (2 A-3 C) \sin (c+d x) \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}{3 d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4087

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[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 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))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3 a A}{2}-\frac{1}{2} a (2 A-3 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{a (2 A-3 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (8 A-3 C)+\frac{9}{4} a^2 C \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=\frac{a^2 (8 A-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}-\frac{a (2 A-3 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{1}{2} (3 a C) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (8 A-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}-\frac{a (2 A-3 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{(3 a 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 )}{d}\\ &=\frac{3 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{a^2 (8 A-3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}-\frac{a (2 A-3 C) \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 6.42657, size = 382, normalized size = 2.26 \[ \frac{6 C \sin (c+d x) \cos ^3(c+d x) \sqrt{\sec ^2(c+d x)-1} (a (\sec (c+d x)+1))^{3/2} \left (\log \left (\sec ^{\frac{3}{2}}(c+d x)+\sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1}+\sqrt{\sec (c+d x)}\right )-\log (\sec (c+d x)+1)\right ) \left (A+C \sec ^2(c+d x)\right )}{d \left (1-\cos ^2(c+d x)\right ) (\sec (c+d x)+1)^{3/2} (A \cos (2 c+2 d x)+A+2 C)}+\frac{(a (\sec (c+d x)+1))^{3/2} \sqrt{(\cos (c+d x)+1) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{2 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (8 A \sin \left (\frac{d x}{2}\right )-3 C \sin \left (\frac{d x}{2}\right )\right )}{3 d}-\frac{2 (8 A-3 C) \tan \left (\frac{c}{2}\right )}{3 d}+\frac{16 A \sin (c) \cos (d x)}{3 d}+\frac{2 A \sin (2 c) \cos (2 d x)}{3 d}+\frac{16 A \cos (c) \sin (d x)}{3 d}+\frac{2 A \cos (2 c) \sin (2 d x)}{3 d}\right )}{\sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)^{3/2} (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(6*C*Cos[c + d*x]^3*(-Log[1 + Sec[c + d*x]] + 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]])*(a*(1 + Sec[c + d*x]))^(3/2)*Sqrt[-1 + Sec[c + d*x]^2]*(A + C*Sec[c + d*x]^2)
*Sin[c + d*x])/(d*(1 - Cos[c + d*x]^2)*(A + 2*C + A*Cos[2*c + 2*d*x])*(1 + Sec[c + d*x])^(3/2)) + (Sqrt[(1 + C
os[c + d*x])*Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2)*(A + C*Sec[c + d*x]^2)*((16*A*Cos[d*x]*Sin[c])/(3*d) +
(2*A*Cos[2*d*x]*Sin[2*c])/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(8*A*Sin[(d*x)/2] - 3*C*Sin[(d*x)/2]))/(3*d)
+ (16*A*Cos[c]*Sin[d*x])/(3*d) + (2*A*Cos[2*c]*Sin[2*d*x])/(3*d) - (2*(8*A - 3*C)*Tan[c/2])/(3*d)))/((A + 2*C
+ A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.362, size = 229, normalized size = 1.4 \begin{align*} -{\frac{a\cos \left ( dx+c \right ) }{12\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 9\,C\cos \left ( dx+c \right ) \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 ) \sqrt{2}-9\,C\cos \left ( dx+c \right ) \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 ) \sqrt{2}+8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+32\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-40\,A\cos \left ( dx+c \right ) +12\,C\cos \left ( dx+c \right ) -12\,C \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(9*C*cos(d*x+c)*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)))*2^(1/2)-9*C*cos(d*x+c)*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)))*2^(1/2)+8*A*cos(d*x+c)^3+32
*A*cos(d*x+c)^2-40*A*cos(d*x+c)+12*C*cos(d*x+c)-12*C)*cos(d*x+c)*(1/cos(d*x+c))^(3/2)/sin(d*x+c)

________________________________________________________________________________________

Maxima [B]  time = 2.05753, size = 1597, normalized size = 9.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 3*(3*(a*log(2*cos(1/2*
d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) +
2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*si
n(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x +
1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)
- a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1
/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2
*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*a*sin(3/2*d*x +
3/2*c) - 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 2*(2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1
/2*c) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(
2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2
*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c
)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x +
2*c) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(
2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2
*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c
)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 4*(sqrt(2)
*a*cos(3/2*d*x + 3/2*c) - sqrt(2)*a*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*C*sqrt(a)/(cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d

________________________________________________________________________________________

Fricas [A]  time = 0.607756, size = 980, normalized size = 5.8 \begin{align*} \left [\frac{9 \,{\left (C a \cos \left (d x + c\right ) + C a\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 (2 \, A a \cos \left (d x + c\right )^{2} + 10 \, A a \cos \left (d x + c\right ) + 3 \, C a\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 )}}}{12 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, \frac{9 \,{\left (C a \cos \left (d x + c\right ) + C a\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 (2 \, A a \cos \left (d x + c\right )^{2} + 10 \, A a \cos \left (d x + c\right ) + 3 \, C a\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 )}}}{6 \,{\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))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(9*(C*a*cos(d*x + c) + C*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*c
os(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*(2*A*a*cos(d*x + c)^2 + 10*A*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/co
s(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/6*(9*(C*a*cos(d*x + c) + C*a)*sqrt(-a)*ar
ctan(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*(2*A*a*cos(d*x + c)^2 + 10*A*a*cos(d*x + c) + 3*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) + d)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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