3.573 \(\int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=317 \[ \frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2+3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}+\frac {2 (a-3 b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d (a-b) (a+b)^{3/2}} \]
[Out]
2/3*(a^2+3*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))
/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^2/(a+b)^(3/2)/d+2/3*(a-3*b)*cot(d*x+c)*EllipticF((a+b*se
c(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2
)/(a-b)/b/(a+b)^(3/2)/d+2/3*a*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*(a^2+3*b^2)*tan(d*x+c)/(a^2-b^
2)^2/d/(a+b*sec(d*x+c))^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.44, antiderivative size = 317, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used =
{3836, 4003, 4005, 3832, 4004} \[ \frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2+3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}+\frac {2 (a-3 b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d (a-b) (a+b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(2*(a^2 + 3*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b
*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*(a - b)*b^2*(a + b)^(3/2)*d) + (2*(a
- 3*b)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec
[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*(a - b)*b*(a + b)^(3/2)*d) + (2*a*Tan[c + d*x
])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*(a^2 + 3*b^2)*Tan[c + d*x])/(3*(a^2 - b^2)^2*d*Sqrt[a + b
*Sec[c + d*x]])
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 3836
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*Cot[e + f*x]*(
a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] - Dist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(b*(m + 1) - a*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1]
Rule 4003
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a*B
)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] &
& LtQ[m, -1]
Rule 4004
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rule 4005
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac {2 a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \int \frac {\sec (c+d x) \left (-\frac {3 b}{2}+\frac {1}{2} a \sec (c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac {2 a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\sec (c+d x) \left (a b+\frac {1}{4} \left (a^2+3 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac {2 a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {(a-3 b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) (a+b)^2}-\frac {\left (a^2+3 b^2\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac {2 \left (a^2+3 b^2\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (a-3 b) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}+\frac {2 a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 13.94, size = 486, normalized size = 1.53 \[ \frac {\sec ^3(c+d x) (a \cos (c+d x)+b)^3 \left (-\frac {2 \left (a^2+3 b^2\right ) \sin (c+d x)}{3 b \left (b^2-a^2\right )^2}+\frac {2 b \sin (c+d x)}{3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac {4 \left (a^2 \sin (c+d x)+b^2 \sin (c+d x)\right )}{3 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}\right )}{d (a+b \sec (c+d x))^{5/2}}+\frac {2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a \cos (c+d x)+b)^2 \left (\left (a^2+3 b^2\right ) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)-2 b \left (a^2+4 a b+3 b^2\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 \left (a^3+a^2 b+3 a b^2+3 b^3\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )}{3 b d \left (a^2-b^2\right )^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^2/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-2*(a^2 + 3*b^2)*Sin[c + d*x])/(3*b*(-a^2 + b^2)^2) + (2*b*Sin[c + d*
x])/(3*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) + (4*(a^2*Sin[c + d*x] + b^2*Sin[c + d*x]))/(3*(-a^2 + b^2)^2*(b +
a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c
+ d*x)/2]^2*Sec[c + d*x]]*(2*(a^3 + a^2*b + 3*a*b^2 + 3*b^3)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a
*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(a^2 +
4*a*b + 3*b^2)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*
EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (a^2 + 3*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c
+ d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*(a^2 - b^2)^2*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2))
________________________________________________________________________________________
fricas [F] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^2/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*
x + c) + a^3), x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate(sec(d*x + c)^2/(b*sec(d*x + c) + a)^(5/2), x)
________________________________________________________________________________________
maple [B] time = 1.26, size = 2411, normalized size = 7.61 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^2/(a+b*sec(d*x+c))^(5/2),x)
[Out]
1/3/d*4^(1/2)*(-4*cos(d*x+c)^2*a^2*b^2+3*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*
cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*b^4+2*cos(d*x+c)^2*
a^3*b+6*cos(d*x+c)^2*a*b^3+cos(d*x+c)*a^2*b^2-sin(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1
/2))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^4-2*cos(d*
x+c)^3*a*b^3-4*cos(d*x+c)*a*b^3-2*cos(d*x+c)^3*a^3*b+cos(d*x+c)^3*a^4-sin(d*x+c)*EllipticE((-1+cos(d*x+c))/sin
(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+
b))^(1/2)*a^4+3*cos(d*x+c)^3*a^2*b^2-3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b
))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)^2*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+
c),((a-b)/(a+b))^(1/2))*a^3*b+4*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1
+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+3*sin(d*x+c)*cos(d
*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+
c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(
d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b-3*sin(d*x+
c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+
cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-3*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^
3-cos(d*x+c)^2*a^4+3*cos(d*x+c)*b^4+3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*b^4+EllipticF((-1+cos(d*x+c))/sin
(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(1/2)*sin(d*x+c)+4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*sin(d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b*sin(d*x+c)-
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(
d*x+c),((a-b)/(a+b))^(1/2))*b^2*a^2*sin(d*x+c)-3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3*a*sin(d*x+c)-3*EllipticE((-1+
cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*b^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))
/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)+sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a^3*b+5*sin(d*x+c)*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d
*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a^2*b^2+7*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))
/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a*b^3-2*sin(
d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c
))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a^3*b-4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*a^2*b^
2-6*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*b^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a-3*cos(d*x+c)^2*b^4)*((b+a*cos(d*x+c))/cos(d*x+c))^
(1/2)/sin(d*x+c)/(b+a*cos(d*x+c))^2/(a-b)^2/(a+b)^2/b
________________________________________________________________________________________
maxima [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)^2/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cos(c + d*x)^2*(a + b/cos(c + d*x))^(5/2)),x)
[Out]
int(1/(cos(c + d*x)^2*(a + b/cos(c + d*x))^(5/2)), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**2/(a+b*sec(d*x+c))**(5/2),x)
[Out]
Integral(sec(c + d*x)**2/(a + b*sec(c + d*x))**(5/2), x)
________________________________________________________________________________________