3.864 \(\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=214 \[ \frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {4 b \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[Out]

2*b^2*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-4*b*(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)/a^2/d/cos(d*
x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2*(a^2-2*b^2)*(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))*cos(d*x+c)^(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)

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Rubi [A]  time = 0.53, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4264, 3847, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {4 b \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(a^2*d*Sqrt[Cos[c + d*x]]*Sqrt
[a + b*Sec[c + d*x]]) + (2*(a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*S
ec[c + d*x]])/(a^2*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*b^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*S
qrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]])

Rule 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 3847

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

Rule 3856

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 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[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 4035

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]

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

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

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Mathematica [C]  time = 9.66, size = 330, normalized size = 1.54 \[ \frac {2 (a \cos (c+d x)+b) \left (a b^2 \sin (c+d x)+\frac {\left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (\left (a^2-2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i a \left (a^2-a b-2 b^2\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+i \left (a^3+a^2 b-2 a b^2-2 b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}\right )}{a^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c))/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.49, size = 997, normalized size = 4.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*Ellip
ticF((-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)*a^2+2*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/s
in(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(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))*a^2+2*cos(d*x+c)*sin(d*x+c)*(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+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*b^2+Elli
pticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+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+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*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)*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*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+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-
cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^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)-2
*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2+a*b*((a-b)/(a+b))^(1/2)+2*b^2*((a-b)/(a+b))^(1/2))/(b+a*cos(d*x+c))/sin(d*
x+c)/((a-b)/(a+b))^(1/2)/(a+b)/a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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