3.7.31 \(\int \frac {\sqrt {a+b \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [631]

Optimal. Leaf size=192 \[ \frac {2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{3 a d \sqrt {a+b \sec (c+d x)}}+\frac {2 b E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

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

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Rubi [A]
time = 0.25, antiderivative size = 192, 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 = {3942, 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 a d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 b \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sec[c + d*x]]/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
*a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*b*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a*d*S
qrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Se
c[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 3942

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

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 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 {\sqrt {a+b \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \int \frac {b+a \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {b \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3 a}+\frac {\left (a^2-b^2\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a}\\ &=\frac {2 \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 a \sqrt {a+b \sec (c+d x)}}+\frac {\left (b \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 a \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 \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 a \sqrt {a+b \sec (c+d x)}}+\frac {\left (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 a \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 a d \sqrt {a+b \sec (c+d x)}}+\frac {2 b E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 \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 = 0.41, size = 156, normalized size = 0.81 \begin {gather*} \frac {2 \sqrt {a+b \sec (c+d x)} \left (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 )+\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 )+a (b+a \cos (c+d x)) \sin (c+d x)\right )}{3 a d (b+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1020\) vs. \(2(228)=456\).
time = 0.27, size = 1021, normalized size = 5.32

method	result	size
default	\(\text {Expression too large to display}\)	\(1021\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(1/2)/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)*(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-
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+cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+c
os(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-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+((b+a*cos(d*x+c))/(1+c
os(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)-((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*Ellipt
icF((-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+a*cos(d*x+c))/(1+c
os(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+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*Ellipt
icE((-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)+cos(d*x+c)^3*((a-b)/(a
+b))^(1/2)*a^2+2*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)-cos(d*x+c)*((a-b)/(a+
b))^(1/2)*a*b+cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2-((a-b)/(a+b))^(1/2)*a*b-((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)/a

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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))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(d*x + c) + a)/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 = 0.83, size = 415, normalized size = 2.16 \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 ) + 3 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 ) - 3 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} + 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} - 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^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(1/2)/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) + 3*I*sqrt(2)*a^(3/2)*b*wei
erstrassZeta(-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)) - 3*I*sqrt(2)*a^(3/2)*b*w
eierstrassZeta(-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 + 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 - 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^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b \sec {\left (c + d x \right )}}}{\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))**(1/2)/sec(d*x+c)**(3/2),x)

[Out]

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

[Out]

integrate(sqrt(b*sec(d*x + c) + a)/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 {\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{{\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))^(1/2)/(1/cos(c + d*x))^(3/2),x)

[Out]

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

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