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

Optimal. Leaf size=237 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}+\frac {b \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}-\frac {\sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.65, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3855, 4109, 3859, 2807, 2805, 3862, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}{d}+\frac {b \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}-\frac {\sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2805

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

Rule 2807

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

Rule 3855

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*d*C
os[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)), x] + Dist[d^2/(2*n - 1), Int[((d
*Csc[e + f*x])^(n - 2)*Simp[2*a*(n - 2) + b*(2*n - 3)*Csc[e + f*x] + a*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] && GtQ[n, 1] && IntegerQ[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 3859

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

Rule 3862

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

Rule 4109

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 5.64, size = 321, normalized size = 1.35 \[ \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {2 a \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{(a+b) \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 i \csc (c+d x) \sqrt {-\frac {a (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {a (\cos (c+d x)+1)}{a-b}} \left (a \left (2 b F\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {b-a}{a+b}\right )+a \Pi \left (1-\frac {a}{b};i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {b-a}{a+b}\right )\right )-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {b-a}{a+b}\right )\right )}{a b \sqrt {\frac {1}{a-b}} \sqrt {a \cos (c+d x)+b}}+4 \tan (c+d x)\right )}{4 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Sec[c + d*x]]*((2*a*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/((a + b)*Sqrt[(b + a*Cos[c + d*x])/
(a + b)]) - ((2*I)*Sqrt[-((a*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(a*(1 + Cos[c + d*x]))/(a - b)]*Csc[c + d*x]*
(-2*b*(a + b)*EllipticE[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*(2*b*Ell
ipticF[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*EllipticPi[1 - a/b, I*Arc
Sinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)])))/(a*Sqrt[(a - b)^(-1)]*b*Sqrt[b + a*Cos
[c + d*x]]) + 4*Tan[c + d*x]))/(4*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.82, size = 789, normalized size = 3.33 \[ -\frac {\left (2 \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \EllipticPi \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \frac {a +b}{a -b}, \frac {i}{\sqrt {\frac {a -b}{a +b}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a -\sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a +\sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) b +2 \EllipticPi \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \frac {a +b}{a -b}, \frac {i}{\sqrt {\frac {a -b}{a +b}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, a -\sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +\sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +\left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}\, a -\cos \left (d x +c \right ) \sqrt {\frac {a -b}{a +b}}\, a +\cos \left (d x +c \right ) \sqrt {\frac {a -b}{a +b}}\, b -\sqrt {\frac {a -b}{a +b}}\, b \right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {\frac {b +a \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}}{d \left (b +a \cos \left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {a -b}{a +b}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-
b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a-((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))*cos(d*x+c)^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)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*b+
2*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(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)*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))*cos(d*x+c)*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)*Ellip
ticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*sin(d*x+c)*b+cos(d*x+c)^2
*((a-b)/(a+b))^(1/2)*a-cos(d*x+c)*((a-b)/(a+b))^(1/2)*a+cos(d*x+c)*((a-b)/(a+b))^(1/2)*b-((a-b)/(a+b))^(1/2)*b
)*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)/((a-b)/(a+b)
)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________