\(\int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [536]

Optimal result

Integrand size = 23, antiderivative size = 277 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {\left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}}-\frac {3 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2881, 3135, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {\left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^2 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {3 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}+\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

-(((a^2 - 3*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(a^2*(a^2 - b^2)*d*Sqrt[(a +
b*Cos[c + d*x])/(a + b)])) + (Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(a*d*S
qrt[a + b*Cos[c + d*x]]) - (3*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/
(a^2*d*Sqrt[a + b*Cos[c + d*x]]) + (b*(a^2 - 3*b^2)*Sin[c + d*x])/(a^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
+ Tan[c + d*x]/(a*d*Sqrt[a + b*Cos[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 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c
+ d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3138

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

Rubi steps \begin{align*} \text {integral}& = \frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}+\frac {\int \frac {\left (-\frac {3 b}{2}+\frac {1}{2} b \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{a} \\ & = \frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}+\frac {2 \int \frac {\left (-\frac {3}{4} b \left (a^2-b^2\right )+\frac {1}{2} a b^2 \cos (c+d x)-\frac {1}{4} b \left (a^2-3 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{a^2 \left (a^2-b^2\right )} \\ & = \frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}-\frac {2 \int \frac {\left (\frac {3}{4} b^2 \left (a^2-b^2\right )-\frac {1}{4} a b \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{a^2 b \left (a^2-b^2\right )}-\frac {\left (a^2-3 b^2\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = \frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}+\frac {\int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{2 a}-\frac {(3 b) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{2 a^2}-\frac {\left (\left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{2 a^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \\ & = -\frac {\left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}}+\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{2 a \sqrt {a+b \cos (c+d x)}}-\frac {\left (3 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{2 a^2 \sqrt {a+b \cos (c+d x)}} \\ & = -\frac {\left (a^2-3 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}}-\frac {3 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a^2 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (a^2-3 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d \sqrt {a+b \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.70 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.59 \[ \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {-\frac {b \left (-\frac {8 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (7 a^2-9 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (a^2-3 b^2\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b^2 \sqrt {-\frac {1}{a+b}}}\right )}{(a-b) (a+b)}+\frac {4 \left (a^3-a b^2+b \left (a^2-3 b^2\right ) \cos (c+d x)\right ) \tan (c+d x)}{\left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{4 a^2 d} \]

[In]

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

[Out]

(-((b*((-8*a*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*
x]] + (2*(7*a^2 - 9*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a
+ b*Cos[c + d*x]] - ((2*I)*(a^2 - 3*b^2)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x])
)/(a - b))]*Csc[c + d*x]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a +
b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*E
llipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b^2*Sqrt[
-(a + b)^(-1)])))/((a - b)*(a + b))) + (4*(a^3 - a*b^2 + b*(a^2 - 3*b^2)*Cos[c + d*x])*Tan[c + d*x])/((a^2 - b
^2)*Sqrt[a + b*Cos[c + d*x]]))/(4*a^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(897\) vs. \(2(348)=696\).

Time = 6.44 (sec) , antiderivative size = 898, normalized size of antiderivative = 3.24

method	result	size
default	\(\text {Expression too large to display}\)	\(898\)

[In]

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

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/a*(-cos(1/2*d*x+1/2*c)/a*(-2*sin(1/2*d*x+1/2
*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*co
s(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(
1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1
/2))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*
b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))+2/a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(
1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos
(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-2*b^2/a^2/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^2)*
(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b+Elli
pticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*a-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b
))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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