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

Optimal. Leaf size=136 \[ \frac{\sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

[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])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - (2*A*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.337145, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {4087, 4013, 3808, 206} \[ \frac{\sqrt{2} (A+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 A \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*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])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - (2*A*Sqrt[Sec[c + d*x]]*Sin[c +
d*x])/(3*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 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{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 A \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{-\frac{a A}{2}+\frac{1}{2} a (2 A+3 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=\frac{2 A \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sqrt{\sec (c+d x)} \sin (c+d x)}{3 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)}{3 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sqrt{\sec (c+d x)} \sin (c+d x)}{3 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)}{3 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 A \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 3.88161, size = 273, normalized size = 2.01 \[ \frac{\sqrt{\sec (c+d x)+1} \left (A+C \sec ^2(c+d x)\right ) \left (3 \sqrt{2} (A+C) \cos ^2(c+d x) \sqrt{\tan ^2(c+d x)} \cot (c+d x) \left (\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )\right )-\frac{16 A \sin ^2\left (\frac{1}{2} (c+d x)\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)+1}}{\sec ^{\frac{5}{2}}(c+d x)}\right )}{6 d \sqrt{a (\sec (c+d x)+1)} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((-16*A*Sqrt[1 + Sec[c + d*x]]*Sin[(c + d*x)/2]^2*Tan[(c + d*x)
/2])/Sec[c + d*x]^(5/2) + 3*Sqrt[2]*(A + C)*Cos[c + d*x]^2*Cot[c + d*x]*(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]] - 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]))/(6*d*(A + 2*C + A*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.378, size = 171, normalized size = 1.3 \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,ad\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 3\,\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 ) +3\,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 ) +2\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,A\cos \left ( dx+c \right ) +2\,A \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+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/3/d/a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(3*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)+3*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)+2*A*cos(d*x+c)^2-4*A*cos(d*x+c)+2*A)*cos(d*x+c)^2*(1/cos(d*x+c))^(3/2)/sin(d*x+c)

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Maxima [B]  time = 2.01865, size = 504, normalized size = 3.71 \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)^(3/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/6*((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*arc
tan2(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(s
in(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*sq
rt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*A/sqrt(a) - 3*(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))*C/sqrt(a))/d

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Fricas [A]  time = 0.5318, size = 917, normalized size = 6.74 \begin{align*} \left [\frac{\frac{3 \, \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 (A \cos \left (d x + c\right )^{2} - A \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 )}}}{6 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac{3 \, \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 (A \cos \left (d x + c\right )^{2} - A \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 )}}}{3 \,{\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)^(3/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*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*(A*cos(d*x + c)^2 - A*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sq
rt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d), -1/3*(3*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*sqrt(-1/a)*ar
ctan(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*cos(d*
x + c)^2 - A*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]  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 )} \sec ^{\frac{3}{2}}{\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)**(3/2)/(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))*sec(c + d*x)**(3/2)), 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} \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+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(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)^(3/2)), x)