3.8.45 \(\int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [745]

Optimal. Leaf size=381 \[ \frac {2 \left (A b^2+a^2 C\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a b^2 \sqrt {a+b} d}-\frac {2 (A b-a C) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a b \sqrt {a+b} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

2*(A*b^2+C*a^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^2/d/(a+b)^(1/2)-2*(A*b-C*a)*cot(d*x+c)*EllipticF((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/d/(a+b)^(1/2)-2*A*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(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^2/d+2*(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2
-b^2)/d/(a+b*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4146, 4143, 4006, 3869, 3917, 4089} \begin {gather*} \frac {2 \left (a^2 C+A 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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a b^2 d \sqrt {a+b}}-\frac {2 A \sqrt {a+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}} \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 (A b-a C) \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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a b d \sqrt {a+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b^2 + a^2*C)*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))])/(a*b^2*Sqrt[a + b]*d) - (2*(A*b - a*C
)*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))])/(a*b*Sqrt[a + b]*d) - (2*A*Sqrt[a + b]*Cot[c + d*x]*El
lipticPi[(a + b)/a, 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))])/(a^2*d) + (2*(A*b^2 + a^2*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d
*Sqrt[a + b*Sec[c + d*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 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 4146

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(
A*b^2 + 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*b*(A + C)*(m + 1)*Csc[e + f*x] +
(A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Int
egerQ[2*m] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {-\frac {1}{2} A \left (a^2-b^2\right )+\frac {1}{2} a b (A+C) \sec (c+d x)+\frac {1}{2} \left (A b^2+a^2 C\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {-\frac {1}{2} A \left (a^2-b^2\right )+\left (\frac {1}{2} a b (A+C)+\frac {1}{2} \left (-A b^2-a^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}-\frac {\left (A b^2+a^2 C\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 \left (A b^2+a^2 C\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}}}{a b^2 \sqrt {a+b} d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {A \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{a}-\frac {(A b-a C) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a (a+b)}\\ &=\frac {2 \left (A b^2+a^2 C\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}}}{a b^2 \sqrt {a+b} d}-\frac {2 (A b-a C) \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}}}{a b \sqrt {a+b} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{a};\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}}}{a^2 d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1119\) vs. \(2(381)=762\).
time = 17.98, size = 1119, normalized size = 2.94 \begin {gather*} \frac {(b+a \cos (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a b \left (-a^2+b^2\right )}+\frac {4 \left (A b^2 \sin (c+d x)+a^2 C \sin (c+d x)\right )}{a \left (a^2-b^2\right ) (b+a \cos (c+d x))}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{3/2}}-\frac {4 (b+a \cos (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a A b^2 \tan \left (\frac {1}{2} (c+d x)\right )+A b^3 \tan \left (\frac {1}{2} (c+d x)\right )+a^3 C \tan \left (\frac {1}{2} (c+d x)\right )+a^2 b C \tan \left (\frac {1}{2} (c+d x)\right )-2 a A b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )-2 a^3 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+a A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )-A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )+a^3 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-a^2 b C \tan ^5\left (\frac {1}{2} (c+d x)\right )+2 a^2 A b \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 A b^3 \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+2 a^2 A b \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 A b^3 \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+(a+b) \left (A b^2+a^2 C\right ) E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-a b (a+b) (A+C) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{a \left (-a^2 b+b^3\right ) d (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((4*(A*b^2 + a^2*C)*Sin[c + d*x])/(a*b*(-a^2 + b^2)) + (4*(A*b^
2*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(a*(a^2 - b^2)*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])
*(a + b*Sec[c + d*x])^(3/2)) - (4*(b + a*Cos[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]
^2)^(-1)]*(a*A*b^2*Tan[(c + d*x)/2] + A*b^3*Tan[(c + d*x)/2] + a^3*C*Tan[(c + d*x)/2] + a^2*b*C*Tan[(c + d*x)/
2] - 2*a*A*b^2*Tan[(c + d*x)/2]^3 - 2*a^3*C*Tan[(c + d*x)/2]^3 + a*A*b^2*Tan[(c + d*x)/2]^5 - A*b^3*Tan[(c + d
*x)/2]^5 + a^3*C*Tan[(c + d*x)/2]^5 - a^2*b*C*Tan[(c + d*x)/2]^5 + 2*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*
x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^
2)/(a + b)] - 2*A*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*S
qrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c +
d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2
+ b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*A*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c +
d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (
a + b)*(A*b^2 + a^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 +
Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - a*b*(a + b)*(A + C)*
EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqr
t[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(a*(-(a^2*b) + b^3)*d*(A + 2*C + A*Cos[2*c
+ 2*d*x])*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c
+ d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2041\) vs. \(2(352)=704\).
time = 0.24, size = 2042, normalized size = 5.36

method	result	size
default	\(\text {Expression too large to display}\)	\(2042\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(C*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b
))^(1/2))*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)*a+A*(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)*sin(d*x+c)*a^2*b+A*(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)*sin(d*x+c)*a*b^2-A*(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)*sin(d*x+c)*a*b^2-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+c
os(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*cos(d*x+c)*sin(d*x+c)*a^
2*b+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)*sin(d*x+c)*a^2*b-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)*sin(d*x
+c)*a^2*b+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)*sin(d*x+c)*a*b^2-A*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))+2*A*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)*Elliptic
Pi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))-C*a^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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))-C*cos(d*x+c
)*a^3+A*cos(d*x+c)^2*a*b^2-A*cos(d*x+c)*a*b^2+C*cos(d*x+c)*a^2*b-A*cos(d*x+c)^2*b^3+C*cos(d*x+c)^2*a^3+A*cos(d
*x+c)*b^3-C*cos(d*x+c)^2*a^2*b-A*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)+2*A*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x
+c),-1,((a-b)/(a+b))^(1/2))*cos(d*x+c)-C*a^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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)+A*a^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)*EllipticF((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*b+A*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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a-A*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/
(a+b))^(1/2))*a-2*A*a^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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b+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))*a^2*b*sin(d
*x+c)-C*a^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)*Ellipti
cE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b)/(b+a*cos(d*x+c))/sin(d*x+c)/a/b/(a+b)/(a-b)

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)/(a + b*sec(c + d*x))**(3/2), 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((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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