3.206 \(\int \frac{c+d \sec (e+f x)}{(a+b \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=495 \[ -\frac{2 \left (6 a^2 b c+a^2 b d-3 a^3 d-a b^2 c-3 b^3 c\right ) \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 a^2 b f (a-b) (a+b)^{3/2}}+\frac{2 b \left (7 a^2 b c-4 a^3 d-3 b^3 c\right ) \tan (e+f x)}{3 a^2 f \left (a^2-b^2\right )^2 \sqrt{a+b \sec (e+f x)}}+\frac{2 b (b c-a d) \tan (e+f x)}{3 a f \left (a^2-b^2\right ) (a+b \sec (e+f x))^{3/2}}+\frac{2 \left (7 a^2 b c-4 a^3 d-3 b^3 c\right ) \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b f (a-b) (a+b)^{3/2}}-\frac{2 c \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^3 f} \]
[Out]
(2*(7*a^2*b*c - 3*b^3*c - 4*a^3*d)*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b
)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(3*a^2*(a - b)*b*(a +
b)^(3/2)*f) - (2*(6*a^2*b*c - a*b^2*c - 3*b^3*c - 3*a^3*d + a^2*b*d)*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b
*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]
))/(a - b))])/(3*a^2*(a - b)*b*(a + b)^(3/2)*f) - (2*Sqrt[a + b]*c*Cot[e + f*x]*EllipticPi[(a + b)/a, ArcSin[S
qrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec
[e + f*x]))/(a - b))])/(a^3*f) + (2*b*(b*c - a*d)*Tan[e + f*x])/(3*a*(a^2 - b^2)*f*(a + b*Sec[e + f*x])^(3/2))
+ (2*b*(7*a^2*b*c - 3*b^3*c - 4*a^3*d)*Tan[e + f*x])/(3*a^2*(a^2 - b^2)^2*f*Sqrt[a + b*Sec[e + f*x]])
________________________________________________________________________________________
Rubi [A] time = 0.777503, antiderivative size = 495, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used =
{3923, 4060, 4058, 3921, 3784, 3832, 4004} \[ \frac{2 b \left (7 a^2 b c-4 a^3 d-3 b^3 c\right ) \tan (e+f x)}{3 a^2 f \left (a^2-b^2\right )^2 \sqrt{a+b \sec (e+f x)}}+\frac{2 b (b c-a d) \tan (e+f x)}{3 a f \left (a^2-b^2\right ) (a+b \sec (e+f x))^{3/2}}-\frac{2 \left (6 a^2 b c+a^2 b d-3 a^3 d-a b^2 c-3 b^3 c\right ) \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b f (a-b) (a+b)^{3/2}}+\frac{2 \left (7 a^2 b c-4 a^3 d-3 b^3 c\right ) \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b f (a-b) (a+b)^{3/2}}-\frac{2 c \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^3 f} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sec[e + f*x])/(a + b*Sec[e + f*x])^(5/2),x]
[Out]
(2*(7*a^2*b*c - 3*b^3*c - 4*a^3*d)*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b
)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(3*a^2*(a - b)*b*(a +
b)^(3/2)*f) - (2*(6*a^2*b*c - a*b^2*c - 3*b^3*c - 3*a^3*d + a^2*b*d)*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b
*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]
))/(a - b))])/(3*a^2*(a - b)*b*(a + b)^(3/2)*f) - (2*Sqrt[a + b]*c*Cot[e + f*x]*EllipticPi[(a + b)/a, ArcSin[S
qrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec
[e + f*x]))/(a - b))])/(a^3*f) + (2*b*(b*c - a*d)*Tan[e + f*x])/(3*a*(a^2 - b^2)*f*(a + b*Sec[e + f*x])^(3/2))
+ (2*b*(7*a^2*b*c - 3*b^3*c - 4*a^3*d)*Tan[e + f*x])/(3*a^2*(a^2 - b^2)^2*f*Sqrt[a + b*Sec[e + f*x]])
Rule 3923
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(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)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
Rule 4060
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(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)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Rule 4058
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, 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, C}, x] && NeQ[a^2 - b^2, 0
]
Rule 3921
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[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]
Rule 3784
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
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 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]
Rubi steps
\begin{align*} \int \frac{c+d \sec (e+f x)}{(a+b \sec (e+f x))^{5/2}} \, dx &=\frac{2 b (b c-a d) \tan (e+f x)}{3 a \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}-\frac{2 \int \frac{-\frac{3}{2} \left (a^2-b^2\right ) c+\frac{3}{2} a (b c-a d) \sec (e+f x)-\frac{1}{2} b (b c-a d) \sec ^2(e+f x)}{(a+b \sec (e+f x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 b (b c-a d) \tan (e+f x)}{3 a \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac{2 b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \tan (e+f x)}{3 a^2 \left (a^2-b^2\right )^2 f \sqrt{a+b \sec (e+f x)}}+\frac{4 \int \frac{\frac{3}{4} \left (a^2-b^2\right )^2 c-\frac{1}{4} a \left (6 a^2 b c-2 b^3 c-3 a^3 d-a b^2 d\right ) \sec (e+f x)-\frac{1}{4} b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \sec ^2(e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 b (b c-a d) \tan (e+f x)}{3 a \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac{2 b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \tan (e+f x)}{3 a^2 \left (a^2-b^2\right )^2 f \sqrt{a+b \sec (e+f x)}}+\frac{4 \int \frac{\frac{3}{4} \left (a^2-b^2\right )^2 c+\left (\frac{1}{4} b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right )-\frac{1}{4} a \left (6 a^2 b c-2 b^3 c-3 a^3 d-a b^2 d\right )\right ) \sec (e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}-\frac{\left (b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right )\right ) \int \frac{\sec (e+f x) (1+\sec (e+f x))}{\sqrt{a+b \sec (e+f x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \cot (e+f x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} f}+\frac{2 b (b c-a d) \tan (e+f x)}{3 a \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac{2 b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \tan (e+f x)}{3 a^2 \left (a^2-b^2\right )^2 f \sqrt{a+b \sec (e+f x)}}+\frac{c \int \frac{1}{\sqrt{a+b \sec (e+f x)}} \, dx}{a^2}+\frac{\left (a b^2 c+3 b^3 c+3 a^3 d-a^2 b (6 c+d)\right ) \int \frac{\sec (e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx}{3 a^2 (a-b) (a+b)^2}\\ &=\frac{2 \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \cot (e+f x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} f}+\frac{2 \left (a b^2 c+3 b^3 c+3 a^3 d-a^2 b (6 c+d)\right ) \cot (e+f x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} f}-\frac{2 \sqrt{a+b} c \cot (e+f x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{a^3 f}+\frac{2 b (b c-a d) \tan (e+f x)}{3 a \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac{2 b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \tan (e+f x)}{3 a^2 \left (a^2-b^2\right )^2 f \sqrt{a+b \sec (e+f x)}}\\ \end{align*}
Mathematica [C] time = 17.1037, size = 2083, normalized size = 4.21 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*Sec[e + f*x])/(a + b*Sec[e + f*x])^(5/2),x]
[Out]
((b + a*Cos[e + f*x])^3*Sec[e + f*x]^2*(c + d*Sec[e + f*x])*((2*(-7*a^2*b*c + 3*b^3*c + 4*a^3*d)*Sin[e + f*x])
/(3*a^2*(a^2 - b^2)^2) - (2*(b^3*c*Sin[e + f*x] - a*b^2*d*Sin[e + f*x]))/(3*a^2*(a^2 - b^2)*(b + a*Cos[e + f*x
])^2) - (2*(-8*a^2*b^2*c*Sin[e + f*x] + 4*b^4*c*Sin[e + f*x] + 5*a^3*b*d*Sin[e + f*x] - a*b^3*d*Sin[e + f*x]))
/(3*a^2*(a^2 - b^2)^2*(b + a*Cos[e + f*x]))))/(f*(d + c*Cos[e + f*x])*(a + b*Sec[e + f*x])^(5/2)) + (2*(b + a*
Cos[e + f*x])^(5/2)*Sec[e + f*x]^(3/2)*(c + d*Sec[e + f*x])*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*
x)/2]^2)/(1 + Tan[(e + f*x)/2]^2)]*(7*a^3*b*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f*x)/2] + 7*a^2*b^2*Sqrt[(-a + b
)/(a + b)]*c*Tan[(e + f*x)/2] - 3*a*b^3*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f*x)/2] - 3*b^4*Sqrt[(-a + b)/(a + b
)]*c*Tan[(e + f*x)/2] - 4*a^4*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2] - 4*a^3*b*Sqrt[(-a + b)/(a + b)]*d*Tan
[(e + f*x)/2] - 14*a^3*b*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f*x)/2]^3 + 6*a*b^3*Sqrt[(-a + b)/(a + b)]*c*Tan[(e
+ f*x)/2]^3 + 8*a^4*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2]^3 + 7*a^3*b*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f
*x)/2]^5 - 7*a^2*b^2*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f*x)/2]^5 - 3*a*b^3*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f
*x)/2]^5 + 3*b^4*Sqrt[(-a + b)/(a + b)]*c*Tan[(e + f*x)/2]^5 - 4*a^4*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2]
^5 + 4*a^3*b*Sqrt[(-a + b)/(a + b)]*d*Tan[(e + f*x)/2]^5 - (6*I)*a^4*c*EllipticPi[-((a + b)/(a - b)), I*ArcSin
h[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[
(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + (12*I)*a^2*b^2*c*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sq
rt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e +
f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - (6*I)*b^4*c*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a +
b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e + f*x)/2]^
2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - (6*I)*a^4*c*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b
)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Tan[(e + f*x)/2]^2*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e
+ f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + (12*I)*a^2*b^2*c*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[
(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Tan[(e + f*x)/2]^2*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a
+ b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - (6*I)*b^4*c*EllipticPi[-((a + b)/(a - b)), I*Arc
Sinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Tan[(e + f*x)/2]^2*Sqrt[1 - Tan[(e + f*x)/2]^2
]*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + I*(a - b)*(-7*a^2*b*c + 3*b^3*c + 4*a^
3*d)*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^
2]*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + I*(a - b)*(-
4*a*b^2*c - 6*b^3*c + 3*a^3*(c - d) + a^2*b*(9*c + d))*EllipticF[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x
)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(a + b - a*Tan[(e + f*x)/2]
^2 + b*Tan[(e + f*x)/2]^2)/(a + b)]))/(3*a^2*Sqrt[(-a + b)/(a + b)]*(a^2 - b^2)^2*f*(d + c*Cos[e + f*x])*(a +
b*Sec[e + f*x])^(5/2)*(-1 + Tan[(e + f*x)/2]^2)*Sqrt[(1 + Tan[(e + f*x)/2]^2)/(1 - Tan[(e + f*x)/2]^2)]*(a*(-1
+ Tan[(e + f*x)/2]^2) - b*(1 + Tan[(e + f*x)/2]^2)))
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Maple [B] time = 0.368, size = 5712, normalized size = 11.5 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sec(f*x+e))/(a+b*sec(f*x+e))^(5/2),x)
[Out]
result too large to display
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d \sec \left (f x + e\right ) + c}{{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+b*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate((d*sec(f*x + e) + c)/(b*sec(f*x + e) + a)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right ) + c\right )}}{b^{3} \sec \left (f x + e\right )^{3} + 3 \, a b^{2} \sec \left (f x + e\right )^{2} + 3 \, a^{2} b \sec \left (f x + e\right ) + a^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+b*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c)/(b^3*sec(f*x + e)^3 + 3*a*b^2*sec(f*x + e)^2 + 3*a^2*b*
sec(f*x + e) + a^3), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c + d \sec{\left (e + f x \right )}}{\left (a + b \sec{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+b*sec(f*x+e))**(5/2),x)
[Out]
Integral((c + d*sec(e + f*x))/(a + b*sec(e + f*x))**(5/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d \sec \left (f x + e\right ) + c}{{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+b*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e) + c)/(b*sec(f*x + e) + a)^(5/2), x)