3.2.71 \(\int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx\) [171]

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

[Out]

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

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Rubi [A]
time = 0.14, 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 = {3989, 3972, 482, 536, 209} \begin {gather*} \frac {2 A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {3 A \text {ArcTan}\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)}} \end {gather*}

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 209

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

Rule 482

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 536

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 3972

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 3989

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])

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) \text {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 \text {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) \text {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) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.55, size = 265, normalized size = 2.28 \begin {gather*} \frac {A \left (\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)}} \left (2 i d x-2 \sinh ^{-1}\left (e^{i (c+d x)}\right )-3 \sqrt {2} \log \left (1-e^{i (c+d x)}\right )-2 \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )+3 \sqrt {2} \log \left (1+e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )-\left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}\right ) \sec ^{\frac {3}{2}}(c+d x) \sin ^3\left (\frac {1}{2} (c+d x)\right )}{d (a-a \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*((Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*((2*I)*d*x - 2*ArcS
inh[E^(I*(c + d*x))] - 3*Sqrt[2]*Log[1 - E^(I*(c + d*x))] - 2*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] + 3*Sqrt[
2]*Log[1 + E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/E^((I/2)*(c + d*x)) - (Cos[(c + d*x)/2]
+ Cos[(3*(c + d*x))/2])*Csc[(c + d*x)/2]^2*Sqrt[Sec[c + d*x]])*Sec[c + d*x]^(3/2)*Sin[(c + d*x)/2]^3)/(d*(a -
a*Sec[c + d*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(99)=198\).
time = 4.63, size = 298, normalized size = 2.57

method	result	size
default	\(-\frac {A \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}\, \cos \left (d x +c \right )+\left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}+3 \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \cos \left (d x +c \right )+4 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )-3 \sqrt {2}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-4 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )\right ) \sqrt {2}}{d \left (\frac {a \left (-1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{3} \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}\)	\(298\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (99) = 198\).
time = 1.49, size = 502, normalized size = 4.33 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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 = 1.07, size = 112, normalized size = 0.97 \begin {gather*} \frac {\frac {3 \, \sqrt {2} A \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}}} - \frac {4 \, A \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}}} - \frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{2 \, d} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {A}{\cos \left (c+d\,x\right )}}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(3/2),x)

[Out]

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(3/2), x)

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