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

Optimal. Leaf size=116 \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac{3 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}-\frac{A \tan (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.199528, antiderivative size = 133, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3904, 3887, 471, 522, 203} \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac{3 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}+\frac{A \sin (c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{2 a d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 6.59493, size = 322, normalized size = 2.78 \[ A \left (\frac{\sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^2(c+d x) \left (-\frac{4 \sin \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{d}+\frac{4 \cos \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{d}-\frac{2 \cot \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{2 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}\right )}{(a-a \sec (c+d x))^{3/2}}-\frac{2 \sqrt{2} e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) \left (\sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac{3 \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{2}}+\tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{d (a-a \sec (c+d x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

A*((-2*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(ArcSinh[E^(I*(c
+ d*x))] - (3*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] + ArcTanh[Sqrt[1
+ E^((2*I)*(c + d*x))]])*Sec[c + d*x]^(3/2)*Sin[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a - a*Sec[c + d*x])
^(3/2)) + (Sec[c + d*x]^2*((4*Cos[c/2]*Cos[(d*x)/2])/d - (2*Cot[c/2]*Csc[c/2 + (d*x)/2])/d + (2*Csc[c/2]*Csc[c
/2 + (d*x)/2]^2*Sin[(d*x)/2])/d - (4*Sin[c/2]*Sin[(d*x)/2])/d)*Sin[c/2 + (d*x)/2]^3)/(a - a*Sec[c + d*x])^(3/2
))

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Maple [B]  time = 0.23, size = 298, normalized size = 2.6 \begin{align*} -{\frac{A\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( \cos \left ( dx+c \right ) \sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{{\frac{3}{2}}}+\sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{{\frac{3}{2}}}+\cos \left ( dx+c \right ) \sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+3\,\cos \left ( dx+c \right ) \sqrt{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +4\,\cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) -\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-3\,\sqrt{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) -4\,\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \right ) \left ({\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \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))/(a-a*sec(d*x+c))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A \sec \left (d x + c\right ) + A}{{\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((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.53126, size = 1285, normalized size = 11.08 \begin{align*} \left [-\frac{3 \, \sqrt{2}{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{-a} \log \left (\frac{2 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{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 )}} +{\left (3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \,{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{-a} \log \left (\frac{2 \,{\left (\cos \left (d x + c\right )^{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 )}} -{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 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 )}}}{4 \,{\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac{3 \, \sqrt{2}{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \,{\left (A \cos \left (d x + c\right ) - A\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 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 )}}}{2 \,{\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(3*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a
*cos(d*x + c) - a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin
(d*x + c) + 4*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x +
c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*(A*cos(d*x + c)^2
+ A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c)), 1/2*(3
*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqr
t(a)*sin(d*x + c)))*sin(d*x + c) - 4*(A*cos(d*x + c) - A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) - a)/cos(d*x + c
))*cos(d*x + c)/(sqrt(a)*sin(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)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(Integral(sec(c + d*x)/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x) + Integ
ral(1/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x))

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Giac [A]  time = 1.87604, size = 262, normalized size = 2.26 \begin{align*} -\frac{A{\left (\frac{3 \, \sqrt{2} \arctan \left (\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{\sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{4 \, \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*A*(3*sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*
sgn(tan(1/2*d*x + 1/2*c))) - 4*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(tan
(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/(a^2*sgn(tan(
1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^2))/d