\(\int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [719]

Optimal result

Integrand size = 23, antiderivative size = 277 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (2 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}+\frac {b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a \left (a^2-b^2\right ) d}-\frac {b \left (5 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3317, 3930, 4187, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac {\left (2 a^2-3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^2 d \left (a^2-b^2\right )}+\frac {b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d \left (a^2-b^2\right )}-\frac {\left (2 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}-\frac {b \left (5 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a-b) (a+b)^2} \]

[In]

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

[Out]

-(((2*a^2 - 3*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*(a^2 - b^2)*d)) + (b*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*(a^2 - b^2)*d) - (b*(5*a^2 - 3*b^2)*Sqrt[C
os[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*(a - b)*(a + b)^2*d) + ((2*a^2
- 3*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a^2*(a^2 - b^2)*d) + (b^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(a*(a^2
- b^2)*d*(b + a*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

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 3317

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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 3934

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

Rule 4187

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

Rule 4191

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(b+a \sec (c+d x))^2} \, dx \\ & = \frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {\int \frac {\sqrt {\sec (c+d x)} \left (\frac {b^2}{2}-a b \sec (c+d x)+\frac {1}{2} \left (2 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{b+a \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2-3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {2 \int \frac {-\frac {1}{4} b \left (2 a^2-3 b^2\right )-\frac {1}{2} a \left (a^2-2 b^2\right ) \sec (c+d x)-\frac {1}{4} b \left (4 a^2-3 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{a^2 \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2-3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {2 \int \frac {-\frac {1}{4} b^2 \left (2 a^2-3 b^2\right )-\left (-\frac {1}{4} a b \left (2 a^2-3 b^2\right )+\frac {1}{2} a b \left (a^2-2 b^2\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^2 b^2 \left (a^2-b^2\right )}-\frac {\left (b \left (5 a^2-3 b^2\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2-3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {b \int \sqrt {\sec (c+d x)} \, dx}{2 a \left (a^2-b^2\right )}-\frac {\left (2 a^2-3 b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\left (b \left (5 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = -\frac {b \left (5 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac {\left (b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a \left (a^2-b^2\right )}-\frac {\left (\left (2 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = -\frac {\left (2 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}+\frac {b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a \left (a^2-b^2\right ) d}-\frac {b \left (5 a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-3 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.78 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 a \left (2 a^2 b-3 b^3+2 a \left (a^2-b^2\right ) \sec (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}}+\frac {\cot (c+d x) \left (-2 a^3 \sec ^{\frac {3}{2}}(c+d x)+3 a b^2 \sec ^{\frac {3}{2}}(c+d x)+2 a^3 \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)-3 a b^2 \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)+2 a \left (2 a^2-3 b^2\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}-2 \left (2 a^3+4 a^2 b-3 a b^2-3 b^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+10 a^2 b \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-6 b^3 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{(a-b) (a+b)}}{2 a^3 d} \]

[In]

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

[Out]

((2*a*(2*a^2*b - 3*b^3 + 2*a*(a^2 - b^2)*Sec[c + d*x])*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])*Sqrt[Se
c[c + d*x]]) + (Cot[c + d*x]*(-2*a^3*Sec[c + d*x]^(3/2) + 3*a*b^2*Sec[c + d*x]^(3/2) + 2*a^3*Cos[2*(c + d*x)]*
Sec[c + d*x]^(3/2) - 3*a*b^2*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) + 2*a*(2*a^2 - 3*b^2)*EllipticE[ArcSin[Sqrt[S
ec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*(2*a^3 + 4*a^2*b - 3*a*b^2 - 3*b^3)*EllipticF[ArcSin[Sqrt[Sec[c +
d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 10*a^2*b*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c +
d*x]^2] - 6*b^3*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/((a - b)*(a + b)))
/(2*a^3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(846\) vs. \(2(339)=678\).

Time = 5.98 (sec) , antiderivative size = 847, normalized size of antiderivative = 3.06

method	result	size
default	\(\text {Expression too large to display}\)	\(847\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/a^2/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2
-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+4*b^2/a^2/(-2*a*b+2
*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2/a*b*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))-1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2
*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1
/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/s
in(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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