\(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [655]

Optimal result

Integrand size = 25, antiderivative size = 206 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{b d \sqrt {a+b \sec (c+d x)}}+\frac {2 a E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{b \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}-\frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3930, 4193, 3944, 2886, 2884, 21, 3941, 2734, 2732} \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{b d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b d \sqrt {a+b \sec (c+d x)}} \]

[In]

Int[Sec[c + d*x]^(5/2)/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 3930

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

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 3944

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d*Sqrt
[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 4193

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.83 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.11 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {(b+a \cos (c+d x))^{3/2} \left (\frac {4 a b \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 i \sqrt {-\frac {a (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {a (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 b (a+b) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \left (2 b \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )+a \operatorname {EllipticPi}\left (1-\frac {a}{b},i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )\right )\right )}{\sqrt {\frac {1}{a-b}} b}\right )}{(a-b) (a+b)}+\frac {4 a^2 (b+a \cos (c+d x)) \sin (c+d x)}{-a^2+b^2}\right )}{2 b d (a+b \sec (c+d x))^{3/2}} \]

[In]

Integrate[Sec[c + d*x]^(5/2)/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(Sec[c + d*x]^(3/2)*(((b + a*Cos[c + d*x])^(3/2)*((4*a*b*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x
)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + (2*(3*a^2 - 2*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elliptic
Pi[2, (c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + ((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*EllipticF[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]
], (-a + b)/(a + b)] + a*EllipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/
(a + b)])))/(Sqrt[(a - b)^(-1)]*b)))/((a - b)*(a + b)) + (4*a^2*(b + a*Cos[c + d*x])*Sin[c + d*x])/(-a^2 + b^2
)))/(2*b*d*(a + b*Sec[c + d*x])^(3/2))

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.57 (sec) , antiderivative size = 1019, normalized size of antiderivative = 4.95

method	result	size
default	\(\text {Expression too large to display}\)	\(1019\)

[In]

int(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/d/((a-b)/(a+b))^(1/2)/b/(a+b)*(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(5/2)*
((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^3*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/((1
-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(((a-b)/(a+b))^(1/2)*a*(1-cos(d*x+c))^3*csc(d*x+c)^3+2*(-(a*(1-cos(d*x+c
))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*El
lipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a+(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^
2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(
a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b-(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c
))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-co
t(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a-2*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c
)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticPi(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(
d*x+c)),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a-2*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c
)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticPi(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(
d*x+c)),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*b+((a-b)/(a+b))^(1/2)*a*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2
*csc(d*x+c)^2+1)^3/(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)

Fricas [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**(5/2)/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^(5/2)/(b*sec(d*x + c) + a)^(3/2), x)

Giac [F]

\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(5/2)/(b*sec(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((1/cos(c + d*x))^(5/2)/(a + b/cos(c + d*x))^(3/2),x)

[Out]

int((1/cos(c + d*x))^(5/2)/(a + b/cos(c + d*x))^(3/2), x)