3.5.65 \(\int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx\) [465]

Optimal. Leaf size=235 \[ -\frac {2 (2 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-2 A b^2+a b B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

2*b*(A*b-B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2*(2*A*b-B*a)*(cos(1/2*d*x+1/2*
c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^
(1/2)*sec(d*x+c)^(1/2)/a^2/d/(a+b*sec(d*x+c))^(1/2)+2*(A*a^2-2*A*b^2+B*a*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1
/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a^2/(a^2-b^2)/d/((b
+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)

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Rubi [A]
time = 0.36, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4115, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {2 b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A+a b B-2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 (2 A b-a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a^2 d \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*A*b - a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])
/(a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(a^2*A - 2*A*b^2 + a*b*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a
+ b*Sec[c + d*x]])/(a^2*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a*B
)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3941

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]]), Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4115

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(
a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4120

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-a^2 A+2 A b^2-a b B\right )+\frac {1}{2} a (A b-a B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {(2 A b-a B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2}+\frac {\left (a^2 A-2 A b^2+a b B\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\left ((2 A b-a B) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{a^2 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (a^2 A-2 A b^2+a b B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {\left ((2 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{a^2 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (a^2 A-2 A b^2+a b B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{a^2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=-\frac {2 (2 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-2 A b^2+a b B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.07, size = 178, normalized size = 0.76 \begin {gather*} -\frac {2 \sqrt {\sec (c+d x)} \left (-\left ((a+b) \left (a^2 A-2 A b^2+a b B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )\right )-\left (a^2-b^2\right ) (-2 A b+a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+a b (-A b+a B) \sin (c+d x)\right )}{a^2 (a-b) (a+b) d \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[Sec[c + d*x]]*(-((a + b)*(a^2*A - 2*A*b^2 + a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticE[(c +
d*x)/2, (2*a)/(a + b)]) - (a^2 - b^2)*(-2*A*b + a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2,
(2*a)/(a + b)] + a*b*(-(A*b) + a*B)*Sin[c + d*x]))/(a^2*(a - b)*(a + b)*d*Sqrt[a + b*Sec[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1447\) vs. \(2(277)=554\).
time = 22.76, size = 1448, normalized size = 6.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1448\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(A*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(
a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*a^2-
2*A*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1
+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2-A*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c)
)/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c
),(-(a+b)/(a-b))^(1/2))*a^2-2*A*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(
d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+B*cos(d*x+c)
*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*(
(a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+B*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b
))^(1/2))*a^2+A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c
))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*sin(d*x+c)-2*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1
/2))*b^2*sin(d*x+c)-A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos
(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*sin(d*x+c)-2*A*((b+a*cos(d*x+c))/(1+cos(d*x+
c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-
b))^(1/2))*a*b*sin(d*x+c)+B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((
-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b*sin(d*x+c)+B*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)
/(a-b))^(1/2))*a^2*sin(d*x+c)+A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2+A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b-A*
((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2+2*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2-B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b
-A*((a-b)/(a+b))^(1/2)*a*b-2*A*((a-b)/(a+b))^(1/2)*b^2+B*((a-b)/(a+b))^(1/2)*a*b)/(1/cos(d*x+c))^(1/2)/(b+a*co
s(d*x+c))/sin(d*x+c)/((a-b)/(a+b))^(1/2)/(a+b)/a^2

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.77, size = 679, normalized size = 2.89 \begin {gather*} -\frac {6 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (3 i \, B a^{3} b - 5 i \, A a^{2} b^{2} - 2 i \, B a b^{3} + 4 i \, A b^{4} + {\left (3 i \, B a^{4} - 5 i \, A a^{3} b - 2 i \, B a^{2} b^{2} + 4 i \, A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + \sqrt {2} {\left (-3 i \, B a^{3} b + 5 i \, A a^{2} b^{2} + 2 i \, B a b^{3} - 4 i \, A b^{4} + {\left (-3 i \, B a^{4} + 5 i \, A a^{3} b + 2 i \, B a^{2} b^{2} - 4 i \, A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) - 3 \, \sqrt {2} {\left (i \, A a^{3} b + i \, B a^{2} b^{2} - 2 i \, A a b^{3} + {\left (i \, A a^{4} + i \, B a^{3} b - 2 i \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) - 3 \, \sqrt {2} {\left (-i \, A a^{3} b - i \, B a^{2} b^{2} + 2 i \, A a b^{3} + {\left (-i \, A a^{4} - i \, B a^{3} b + 2 i \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right )}{3 \, {\left ({\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b - a^{3} b^{3}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/3*(6*(B*a^3*b - A*a^2*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(2
)*(3*I*B*a^3*b - 5*I*A*a^2*b^2 - 2*I*B*a*b^3 + 4*I*A*b^4 + (3*I*B*a^4 - 5*I*A*a^3*b - 2*I*B*a^2*b^2 + 4*I*A*a*
b^3)*cos(d*x + c))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*
cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt(2)*(-3*I*B*a^3*b + 5*I*A*a^2*b^2 + 2*I*B*a*b^3 - 4*I*A*b^4
+ (-3*I*B*a^4 + 5*I*A*a^3*b + 2*I*B*a^2*b^2 - 4*I*A*a*b^3)*cos(d*x + c))*sqrt(a)*weierstrassPInverse(-4/3*(3*a
^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*sqrt(2)*
(I*A*a^3*b + I*B*a^2*b^2 - 2*I*A*a*b^3 + (I*A*a^4 + I*B*a^3*b - 2*I*A*a^2*b^2)*cos(d*x + c))*sqrt(a)*weierstra
ssZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/2
7*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) - 3*sqrt(2)*(-I*A*a^3*b - I*B*a
^2*b^2 + 2*I*A*a*b^3 + (-I*A*a^4 - I*B*a^3*b + 2*I*A*a^2*b^2)*cos(d*x + c))*sqrt(a)*weierstrassZeta(-4/3*(3*a^
2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^
3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)))/((a^6 - a^4*b^2)*d*cos(d*x + c) + (a^5*b - a^3*
b^3)*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\sec {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(3/2)/sec(d*x+c)**(1/2),x)

[Out]

Integral((A + B*sec(c + d*x))/((a + b*sec(c + d*x))**(3/2)*sqrt(sec(c + d*x))), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(1/2)),x)

[Out]

int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(1/2)), x)

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