3.3.6 \(\int \frac {c+d \sec (e+f x)}{(a+b \sec (e+f x))^{5/2}} \, dx\) [206]

Optimal. Leaf size=495 \[ \frac {2 \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \cot (e+f x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} f}-\frac {2 \left (6 a^2 b c-a b^2 c-3 b^3 c-3 a^3 d+a^2 b d\right ) \cot (e+f x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} f}-\frac {2 \sqrt {a+b} c \cot (e+f x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{a^3 f}+\frac {2 b (b c-a d) \tan (e+f x)}{3 a \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2}}+\frac {2 b \left (7 a^2 b c-3 b^3 c-4 a^3 d\right ) \tan (e+f x)}{3 a^2 \left (a^2-b^2\right )^2 f \sqrt {a+b \sec (e+f x)}} \]

[Out]

2/3*(-4*a^3*d+7*a^2*b*c-3*b^3*c)*cot(f*x+e)*EllipticE((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*
(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/a^2/(a-b)/b/(a+b)^(3/2)/f-2/3*(-3*a^3*d+6*a^2*b
*c+a^2*b*d-a*b^2*c-3*b^3*c)*cot(f*x+e)*EllipticF((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1
-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/a^2/(a-b)/b/(a+b)^(3/2)/f-2*c*cot(f*x+e)*EllipticPi(
(a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b
*(1+sec(f*x+e))/(a-b))^(1/2)/a^3/f+2/3*b*(-a*d+b*c)*tan(f*x+e)/a/(a^2-b^2)/f/(a+b*sec(f*x+e))^(3/2)+2/3*b*(-4*
a^3*d+7*a^2*b*c-3*b^3*c)*tan(f*x+e)/a^2/(a^2-b^2)^2/f/(a+b*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.54, antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4008, 4145, 4143, 4006, 3869, 3917, 4089} \begin {gather*} -\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};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^3 f}+\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 (-4 a^3 d+7 a^2 b 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}} E\left (\text {ArcSin}\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 (-3 a^3 d+6 a^2 b c+a^2 b 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 (\text {ArcSin}\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 (-4 a^3 d+7 a^2 b c-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)}} \end {gather*}

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 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*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)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(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)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4006

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 4008

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4143

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 4145

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]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 16.79, size = 2083, normalized size = 4.21 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[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] Leaf count of result is larger than twice the leaf count of optimal. \(5709\) vs. \(2(456)=912\).
time = 3.25, size = 5710, normalized size = 11.54

method	result	size
default	\(\text {Expression too large to display}\)	\(5710\)

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,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {c+\frac {d}{\cos \left (e+f\,x\right )}}{{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/cos(e + f*x))/(a + b/cos(e + f*x))^(5/2),x)

[Out]

int((c + d/cos(e + f*x))/(a + b/cos(e + f*x))^(5/2), x)

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