\(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [620]

Optimal result

Integrand size = 23, antiderivative size = 304 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {b \left (4 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \]

[Out]

1/3*(2*a^2-5*b^2)*sin(d*x+c)/a^2/(a^2-b^2)/d/sec(d*x+c)^(1/2)+b^2*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))/se
c(d*x+c)^(1/2)-b*(4*a^2-5*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^3/(a^2-b^2)/d+1/3*(2*a^4+16*a^2*b^2-15*b^4)*(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^4/(a^2-b^2)
/d-b^3*(7*a^2-5*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*a/(a+b),2
^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^4/(a-b)/(a+b)^2/d

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3932, 4189, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^4 d \left (a^2-b^2\right )}-\frac {b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^4 d (a-b) (a+b)^2}-\frac {b \left (4 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )} \]

[In]

Int[1/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

-((b*(4*a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a^2 - b^2)*d)) + (
(2*a^4 + 16*a^2*b^2 - 15*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b
^2)*d) - (b^3*(7*a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])
/(a^4*(a - b)*(a + b)^2*d) + ((2*a^2 - 5*b^2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]) + (b^2*Si
n[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*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 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 3932

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b^2*Co
t[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 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 4189

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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*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] && LeQ[n, -1]

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}& = \frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\int \frac {-a^2+\frac {5 b^2}{2}+a b \sec (c+d x)-\frac {3}{2} b^2 \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {2 \int \frac {-\frac {3}{4} b \left (4 a^2-5 b^2\right )+\frac {1}{2} a \left (a^2+2 b^2\right ) \sec (c+d x)+\frac {1}{4} b \left (2 a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}+\frac {2 \int \frac {-\frac {3}{4} a b \left (4 a^2-5 b^2\right )-\left (-\frac {3}{4} b^2 \left (4 a^2-5 b^2\right )-\frac {1}{2} a^2 \left (a^2+2 b^2\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}-\frac {\left (b^3 \left (7 a^2-5 b^2\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )} \\ & = \frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\left (b \left (4 a^2-5 b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}-\frac {\left (b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^4 \left (a^2-b^2\right )} \\ & = -\frac {b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}-\frac {\left (b \left (4 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac {\left (\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^4 \left (a^2-b^2\right )} \\ & = -\frac {b \left (4 a^2-5 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^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^4+16 a^2 b^2-15 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b^3 \left (7 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.51 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {\frac {a^2 \left (2 a^2 b-5 b^3+2 a \left (a^2-b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x)) \sqrt {\sec (c+d x)}}+\frac {\cot (c+d x) \left (3 a b \left (4 a^2-5 b^2\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+a \left (2 a^3-12 a^2 b-5 a b^2+15 b^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-3 b \left (a \left (4 a^2-5 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin ^2(c+d x)+b \left (-7 a^2+5 b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )\right )}{(a-b) (a+b)}}{3 a^4 d} \]

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

((a^2*(2*a^2*b - 5*b^3 + 2*a*(a^2 - b^2)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])*Sqrt[Se
c[c + d*x]]) + (Cot[c + d*x]*(3*a*b*(4*a^2 - 5*b^2)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*
x]^2] + a*(2*a^3 - 12*a^2*b - 5*a*b^2 + 15*b^3)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2
] - 3*b*(a*(4*a^2 - 5*b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x]^2 + b*(-7*a^2 + 5*b^2)*EllipticPi[-(b/a), ArcSin[Sq
rt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2])))/((a - b)*(a + b)))/(3*a^4*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1063\) vs. \(2(362)=724\).

Time = 16.94 (sec) , antiderivative size = 1064, normalized size of antiderivative = 3.50

method	result	size
default	\(\text {Expression too large to display}\)	\(1064\)

[In]

int(1/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^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)*(4/3/a^2*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
-cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(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)-4/a^3*(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(co
s(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2/a^4*b^4*(a^2/b/(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*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/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)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+2*(a^2
+2*a*b+3*b^2)/a^4*(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))+8/a^3*b^3/(a^2-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)*EllipticPi(cos(1/2*d*x
+1/2*c),2*a/(a-b),2^(1/2)))/sin(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. \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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