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

Optimal. Leaf size=135 \[ -\frac{\sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

(2*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) - (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*Sqrt[Sec[c + d*x]]*S
in[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.363927, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4087, 4023, 3808, 206, 3801, 215} \[ -\frac{\sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) - (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*Sqrt[Sec[c + d*x]]*S
in[c + d*x])/(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 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{A+C \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 A \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sqrt{\sec (c+d x)} \left (-\frac{a A}{2}+\frac{1}{2} a C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{a}\\ &=\frac{2 A \sqrt{\sec (c+d x)} \sin (c+d x)}{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{C \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{a}\\ &=\frac{2 A \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\frac{(2 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 d}+\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{2 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} 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 \sqrt{\sec (c+d x)} \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 2.97222, size = 504, normalized size = 3.73 \[ -\frac{\tan (c+d x) \left (\frac{8 A}{\sqrt{\frac{1}{\cos (c+d x)+1}}}-8 A \sqrt{\sec (c+d x)} \sqrt{\sec (c+d x)+1}+\sqrt{2} A \sqrt{\tan ^2(c+d x)} \log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )-\sqrt{2} A \sqrt{\tan ^2(c+d x)} \log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )-8 C \sqrt{\tan ^2(c+d x)} \log \left (\sec ^{\frac{3}{2}}(c+d x)+\sqrt{\sec (c+d x)}+\sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1}\right )+\sqrt{2} C \sqrt{\tan ^2(c+d x)} \log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )-\sqrt{2} C \sqrt{\tan ^2(c+d x)} \log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )+8 C \sqrt{\tan ^2(c+d x)} \log (\sec (c+d x)+1)\right )}{4 d (\sec (c+d x)-1) \sqrt{\sec (c+d x)+1} \sqrt{a (\sec (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Tan[c + d*x]*((8*A)/Sqrt[(1 + Cos[c + d*x])^(-1)] - 8*A*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]] + 8*C*Log[
1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2] - 8*C*Log[Sqrt[Sec[c + d*x]] + Sec[c + d*x]^(3/2) + Sqrt[1 + Sec[c + d*
x]]*Sqrt[Tan[c + d*x]^2]]*Sqrt[Tan[c + d*x]^2] + Sqrt[2]*A*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[Tan[c + d*x]^2]]*Sqrt[Tan[c + d*x]^2] + Sqrt[2]*C*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[Tan[c + d*x]^2]]*Sq
rt[Tan[c + d*x]^2] - Sqrt[2]*A*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[Tan[c + d*x]^2]]*Sqrt[Tan[c + d*x]^2] - Sqrt[2]*C*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[Tan[c + d*x]^2]]*Sqrt[Tan[c + d*x]^2]))/(4*d*(-
1 + Sec[c + d*x])*Sqrt[1 + Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.35, size = 273, normalized size = 2. \begin{align*} -{\frac{1}{2\,ad\sin \left ( dx+c \right ) } \left ( -C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sin \left ( dx+c \right ) +C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sin \left ( dx+c \right ) -2\,\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 ) -2\,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 ) +4\,A\cos \left ( dx+c \right ) -4\,A \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}}} \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)^(1/2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/2/d/a*(-C*(-2/(cos(d*x+c)+1))^(1/2)*2^(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)))*sin(d*x+c)+C*(-2/(cos(d*x+c)+1))^(1/2)*2^(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)))*sin(d*x+c)-2*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*A*si
n(d*x+c)-2*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)+4*A*cos(d*x
+c)-4*A)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/(1/cos(d*x+c))^(1/2)

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Maxima [B]  time = 2.10825, size = 783, normalized size = 5.8 \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)^(1/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/2*((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*sqrt(2)*sin(1/2*d*x + 1/2*c
))*A/sqrt(a) + (sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos
(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c)
, cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x
+ c))) + 1) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x +
c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos
(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(
d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c)
, cos(d*x + c))) + 2) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c),
cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x
+ c), cos(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x
+ c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(si
n(d*x + c), cos(d*x + c))) + 2))*C/sqrt(a))/d

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Fricas [A]  time = 0.660838, size = 1351, normalized size = 10.01 \begin{align*} \left [\frac{4 \, 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 ) +{\left (C \cos \left (d x + c\right ) + C\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{\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}}}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac{\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 ) + 2 \, 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 ) +{\left (C \cos \left (d x + c\right ) + C\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 )}{a d \cos \left (d x + c\right ) + a d}\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)^(1/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/2*(4*A*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (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)) + 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))/(a*d
*cos(d*x + c) + a*d), (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*A*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sqrt(cos(d*x + c))*sin(d*x + c) + (C*cos(d*x + c) + C)*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)))/(a*d*cos(d*x + c)
+ a*d)]

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

[Out]

Integral((A + C*sec(c + d*x)**2)/(sqrt(a*(sec(c + d*x) + 1))*sqrt(sec(c + d*x))), x)

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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} \sqrt{\sec \left (d x + c\right )}}\,{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)^(1/2)/(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)*sqrt(sec(d*x + c))), x)