3.283 \(\int \frac{A+C \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=224 \[ -\frac{2 (43 A+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{\sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A + 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])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) - (2*A*Sin[c + d*x])/(35*d*Sec[c
+ d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*(31*A + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*S
ec[c + d*x]]) - (2*(43*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.67462, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4087, 4022, 4013, 3808, 206} \[ -\frac{2 (43 A+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{\sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]),x]

[Out]

(Sqrt[2]*(A + 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])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) - (2*A*Sin[c + d*x])/(35*d*Sec[c
+ d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*(31*A + 35*C)*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*S
ec[c + d*x]]) - (2*(43*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*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 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+C \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{-\frac{a A}{2}+\frac{1}{2} a (6 A+7 C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{7 a}\\ &=\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} a^2 (31 A+35 C)-a^2 A \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 \int \frac{-\frac{1}{8} a^3 (43 A+35 C)+\frac{1}{4} a^3 (31 A+35 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+(A+C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 A \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}-\frac{(2 (A+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+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)}{7 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 (31 A+35 C) \sin (c+d x)}{105 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 6.47997, size = 573, normalized size = 2.56 \[ \frac{(A+C) \sin (c+d x) \cos ^4(c+d x) (\sec (c+d x)+1)^{3/2} \sqrt{\sec ^2(c+d x)-1} \left (\log \left (-3 \sec ^2(c+d x)-2 \sqrt{2} \sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1} \sqrt{\sec (c+d x)}-2 \sec (c+d x)+1\right )-\log \left (-3 \sec ^2(c+d x)+2 \sqrt{2} \sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1} \sqrt{\sec (c+d x)}-2 \sec (c+d x)+1\right )\right ) \left (A+C \sec ^2(c+d x)\right )}{d (\cos (c+d x)+1) \sqrt{2-2 \cos ^2(c+d x)} \sqrt{1-\cos ^2(c+d x)} \sqrt{a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)}+\frac{\sqrt{\sec (c+d x)+1} \sqrt{(\cos (c+d x)+1) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{2 (193 A+140 C) \sin (c) \cos (d x)}{105 d}+\frac{(113 A+70 C) \sin (2 c) \cos (2 d x)}{105 d}-\frac{2 (193 A+140 C) \cos (c) \sin (d x)}{105 d}+\frac{(113 A+70 C) \cos (2 c) \sin (2 d x)}{105 d}+\frac{8 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (46 A \sin \left (\frac{d x}{2}\right )+35 C \sin \left (\frac{d x}{2}\right )\right )}{105 d}+\frac{8 (46 A+35 C) \tan \left (\frac{c}{2}\right )}{105 d}-\frac{6 A \sin (3 c) \cos (3 d x)}{35 d}+\frac{A \sin (4 c) \cos (4 d x)}{14 d}-\frac{6 A \cos (3 c) \sin (3 d x)}{35 d}+\frac{A \cos (4 c) \sin (4 d x)}{14 d}\right )}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]),x]

[Out]

((A + C)*Cos[c + d*x]^4*(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])^(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])*Sqrt[2 - 2*Cos[c + d*x]^2]*Sqrt[1 - Cos[c + d*x]^2]*(A +
2*C + A*Cos[2*c + 2*d*x])*Sqrt[a*(1 + Sec[c + d*x])]) + (Sqrt[(1 + Cos[c + d*x])*Sec[c + d*x]]*Sqrt[1 + Sec[c
+ d*x]]*(A + C*Sec[c + d*x]^2)*((-2*(193*A + 140*C)*Cos[d*x]*Sin[c])/(105*d) + ((113*A + 70*C)*Cos[2*d*x]*Sin
[2*c])/(105*d) - (6*A*Cos[3*d*x]*Sin[3*c])/(35*d) + (A*Cos[4*d*x]*Sin[4*c])/(14*d) + (8*Sec[c/2]*Sec[c/2 + (d*
x)/2]*(46*A*Sin[(d*x)/2] + 35*C*Sin[(d*x)/2]))/(105*d) - (2*(193*A + 140*C)*Cos[c]*Sin[d*x])/(105*d) + ((113*A
+ 70*C)*Cos[2*c]*Sin[2*d*x])/(105*d) - (6*A*Cos[3*c]*Sin[3*d*x])/(35*d) + (A*Cos[4*c]*Sin[4*d*x])/(14*d) + (8
*(46*A + 35*C)*Tan[c/2])/(105*d)))/((A + 2*C + 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 = 0.431, size = 216, normalized size = 1. \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{105\,ad\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 30\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-36\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+105\,\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 ) +105\,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 ) +68\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+70\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}-148\,A\cos \left ( dx+c \right ) -140\,C\cos \left ( dx+c \right ) +86\,A+70\,C \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/105/d/a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(30*A*cos(d*x+c)^4-36*A*cos(d*x+c)^3+105*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)+105*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*s
in(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+68*A*cos(d*x+c)^2+70*C*cos(d*x+c)^2-148*A*cos(d*x+c)-140*C*cos
(d*x+c)+86*A+70*C)*cos(d*x+c)^4*(1/cos(d*x+c))^(7/2)/sin(d*x+c)

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Maxima [B]  time = 2.18826, size = 986, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/840*(sqrt(2)*(525*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 175*c
os(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*cos(2/7*arctan2(sin(7/2*
d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 525*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d
*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d
*x + 7/2*c))) - 21*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 420*log
(cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/
2*d*x + 7/2*c)))^2 + 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 1) + 420*log(cos(1/7*arc
tan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*
c)))^2 - 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 1) - 30*sin(7/2*d*x + 7/2*c) + 21*si
n(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 175*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c))) + 525*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A/sqrt(a) + 140*(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), cos(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*arc
tan2(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*ar
ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*C/sqrt(a))/d

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Fricas [A]  time = 0.54735, size = 1108, normalized size = 4.95 \begin{align*} \left [\frac{\frac{105 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right ) +{\left (A + 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 (15 \, A \cos \left (d x + c\right )^{4} - 3 \, A \cos \left (d x + c\right )^{3} +{\left (31 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (43 \, A + 35 \, 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 )}}}{210 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac{105 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right ) +{\left (A + 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 (15 \, A \cos \left (d x + c\right )^{4} - 3 \, A \cos \left (d x + c\right )^{3} +{\left (31 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (43 \, A + 35 \, 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 )}}}{105 \,{\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+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/210*(105*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + 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*cos(d*x
+ c) + 1))/sqrt(a) + 4*(15*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 + (31*A + 35*C)*cos(d*x + c)^2 - (43*A + 35*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), -1/105*(105*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + 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*(15*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 +
(31*A + 35*C)*cos(d*x + c)^2 - (43*A + 35*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)]

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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+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2)/(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} + A}{\sqrt{a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(sqrt(a*sec(d*x + c) + a)*sec(d*x + c)^(7/2)), x)