3.7.37 \(\int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [637]

Optimal. Leaf size=187 \[ \frac {2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{3 d \sqrt {a+b \sec (c+d x)}}+\frac {8 b E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/3*(a^2-b^2)*(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)/d/(a+b*sec(d*x+c))^(1/2)+2/3*a*sin(d*x+c)*(a+b*sec(d*x+c
))^(1/2)/d/sec(d*x+c)^(1/2)+8/3*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)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3949, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {2 \left (a^2-b^2\right ) \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 )}{3 d \sqrt {a+b \sec (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\sec (c+d x)}}+\frac {8 b \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3
*d*Sqrt[a + b*Sec[c + d*x]]) + (8*b*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*d*Sqrt[
(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[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 3949

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2), x_Symbol] :> Simp[a*Cot
[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n +
1)/Sqrt[a + b*Csc[e + f*x]])*Simp[a*b*(2*n - 1) + 2*(b^2*n + a^2*(n + 1))*Csc[e + f*x] + a*b*(2*n + 3)*Csc[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegersQ[2*n]

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))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {1}{3} \int \frac {-4 a b-\left (a^2+3 b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} (4 b) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (-a^2+b^2\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {\left (\left (-a^2+b^2\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 b \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {\left (\left (-a^2+b^2\right ) \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}{3 \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 b \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{3 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=\frac {2 \left (a^2-b^2\right ) \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)}}{3 d \sqrt {a+b \sec (c+d x)}}+\frac {8 b E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.46, size = 156, normalized size = 0.83 \begin {gather*} \frac {(a+b \sec (c+d x))^{3/2} \left (8 b (a+b) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+2 a (b+a \cos (c+d x)) \sin (c+d x)\right )}{3 d (b+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sec[c + d*x])^(3/2)*(8*b*(a + b)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*a)/(a +
b)] + 2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + 2*a*(b + a*Cos[
c + d*x])*Sin[c + d*x]))/(3*d*(b + a*Cos[c + d*x])^2*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. \(1218\) vs. \(2(223)=446\).
time = 0.25, size = 1219, normalized size = 6.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(4*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-4*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+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^2-4*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+3*((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)*EllipticF((-1+cos(d*x+c))*((a-b
)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2+cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2+4*((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)-4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*E
llipticE((-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)+((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)-4*((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)+3*((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))*b^2*sin(d*x+c)+5*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b-((a-b)/(a+b))^(1/2)*a^2*cos(d
*x+c)-4*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b+4*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2-((a-b)/(a+b))^(1/2)*a*b-4*((a-
b)/(a+b))^(1/2)*b^2)*cos(d*x+c)^2*(1/cos(d*x+c))^(3/2)/sin(d*x+c)/(b+a*cos(d*x+c))/((a-b)/(a+b))^(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+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.94, size = 415, normalized size = 2.22 \begin {gather*} \frac {6 \, a^{2} \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 ) + 12 i \, \sqrt {2} a^{\frac {3}{2}} b {\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 ) - 12 i \, \sqrt {2} a^{\frac {3}{2}} b {\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 ) + \sqrt {2} {\left (-3 i \, a^{2} - i \, b^{2}\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 \, a^{2} + i \, b^{2}\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 )}{9 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(6*a^2*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 12*I*sqrt(2)*a^(3/2)*b*we
ierstrassZeta(-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)) - 12*I*sqrt(2)*a^(3/2)*b
*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)) + sqrt(2)*(-3*I*a^2 -
I*b^2)*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*a^2 + I*b^2)*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))/(a*d)

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

[Out]

Integral((a + b*sec(c + d*x))**(3/2)/sec(c + d*x)**(3/2), 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))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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

[Out]

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

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