3.746 \(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=431 \[ \frac {3 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^3 d}-\frac {b \left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\left (2 a^2 C+a A b+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 b d \sqrt {a+b}}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 b d \sqrt {a+b}}+\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}} \]
[Out]
-(3*A*b^2-a^2*(A-2*C))*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^2/b/d/(a+b)^(1/2)+(A*a*b+3*A*b^2+2*C*a^2)*cot(d*x+c)*El
lipticF((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^2/b/d/(a+b)^(1/2)+3*A*b*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^3/d+A*sin(d*x+
c)/a/d/(a+b*sec(d*x+c))^(1/2)-b*(3*A*b^2-a^2*(A-2*C))*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
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Rubi [A] time = 0.66, antiderivative size = 431, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used =
{4105, 4060, 4058, 3921, 3784, 3832, 4004} \[ -\frac {b \left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\left (2 a^2 C+a A b+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 b d \sqrt {a+b}}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 b d \sqrt {a+b}}+\frac {3 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^3 d}+\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(3/2),x]
[Out]
-(((3*A*b^2 - a^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^2*b*Sqrt[a + b]*d)) + ((
a*A*b + 3*A*b^2 + 2*a^2*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^2*b*Sqrt[a + b]*d) + (3*A*
b*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(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^3*d) + (A*Sin[c + d*x])/(a*
d*Sqrt[a + b*Sec[c + d*x]]) - (b*(3*A*b^2 - a^2*(A - 2*C))*Tan[c + d*x])/(a^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c +
d*x]])
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 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 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 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 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 4105
Int[((A_.) + 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*(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, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]
Rubi steps
\begin {align*} \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx &=\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\frac {3 A b}{2}-a C \sec (c+d x)-\frac {1}{2} A b \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{a}\\ &=\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {b \left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \int \frac {-\frac {3}{4} A b \left (a^2-b^2\right )+\frac {1}{2} a \left (A b^2+a^2 C\right ) \sec (c+d x)+\frac {1}{4} b \left (3 A b^2-a^2 (A-2 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {b \left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \int \frac {-\frac {3}{4} A b \left (a^2-b^2\right )+\left (-\frac {1}{4} b \left (3 A b^2-a^2 (A-2 C)\right )+\frac {1}{2} a \left (A b^2+a^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2 \left (a^2-b^2\right )}+\frac {\left (b \left (3 A b^2-a^2 (A-2 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 A b^2-a^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^2 b \sqrt {a+b} d}+\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {b \left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {(3 A b) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^2}+\frac {\left (a A b+3 A b^2+2 a^2 C\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^2 (a+b)}\\ &=-\frac {\left (3 A b^2-a^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^2 b \sqrt {a+b} d}+\frac {\left (a A b+3 A b^2+2 a^2 C\right ) \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^2 b \sqrt {a+b} d}+\frac {3 A b \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^3 d}+\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {b \left (3 A b^2-a^2 (A-2 C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [B] time = 19.98, size = 1251, normalized size = 2.90 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]*(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^2*(a^2 - b^2)) - (4*(A*b^3
*Sin[c + d*x] + a^2*b*C*Sin[c + d*x]))/(a^2*(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)) - (2*(b + a*Cos[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)
/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(a^3*A*Tan[(
c + d*x)/2] + a^2*A*b*Tan[(c + d*x)/2] - 3*a*A*b^2*Tan[(c + d*x)/2] - 3*A*b^3*Tan[(c + d*x)/2] - 2*a^3*C*Tan[(
c + d*x)/2] - 2*a^2*b*C*Tan[(c + d*x)/2] - 2*a^3*A*Tan[(c + d*x)/2]^3 + 6*a*A*b^2*Tan[(c + d*x)/2]^3 + 4*a^3*C
*Tan[(c + d*x)/2]^3 + a^3*A*Tan[(c + d*x)/2]^5 - a^2*A*b*Tan[(c + d*x)/2]^5 - 3*a*A*b^2*Tan[(c + d*x)/2]^5 + 3
*A*b^3*Tan[(c + d*x)/2]^5 - 2*a^3*C*Tan[(c + d*x)/2]^5 + 2*a^2*b*C*Tan[(c + d*x)/2]^5 - 6*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)] + 6*A*b^3*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)] - 6*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)] + 6*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)*(-3*A*b^2 + a^2*(A - 2*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sq
rt[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)] + 2*a*(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)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(a^2*(a
^2 - b^2)*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*Sqrt[1 + Tan[(c + d*x
)/2]^2]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))
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fricas [F] time = 23.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)*sec(d*x + c)^2 + A*cos(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b
*sec(d*x + c) + a^2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(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)*cos(d*x + c)/(b*sec(d*x + c) + a)^(3/2), x)
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maple [B] time = 1.75, size = 2489, normalized size = 5.77 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x)
[Out]
-1/2/d*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(A*cos(d*x+c)^3*a^3-A*cos(d*x+c)^2*a^3+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))*a^3*sin(d*x+c)-3*A*cos(d*x+c)^2*b^3-A*cos(d*x+c)^3*a*b^2+A*cos(d*x+c)^2*a^2*b+3*A*cos(d*x+c)^2*a*b^2
-A*cos(d*x+c)*a^2*b-2*A*cos(d*x+c)*a*b^2+2*C*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*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)*a^3+3*A*cos(d*x+c)*b^3-3*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))*b^3*sin(d*x+c)-2*C*cos(d*x+c)*a^3+2*C*cos(d*x+c)^2*a^3-2*C*cos(d*x+c)^2*a^2*b+2*C
*cos(d*x+c)*a^2*b+6*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)*Elliptic
Pi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)-2*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*sin(d
*x+c)-2*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)*cos(d
*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b+2*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)*cos(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/
(a+b))^(1/2))*b-6*A*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*cos(d*x+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)*b+2*A*EllipticF((-1+cos(d*x+c))/
sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+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)*b+2*A*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*b^2*(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)*sin(d*x+c)*a+A*EllipticE((-1+cos(
d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+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)*b-3*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))*sin(d*x+c)*cos(d*x+c)*a*b^2-6*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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),
-1,((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)+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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)+2*A*(cos(d*x+c)/(1+c
os(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^2*sin(d*x+c)+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))*a^2*b*sin(d*x+c)-3*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))*a*
b^2*sin(d*x+c)-2*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b+2*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*co
s(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)+6*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)*EllipticPi((-1+cos(d*x+
c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^3+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))*sin(d*x+c)*cos(
d*x+c)*a^3-3*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+c
os(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^3-2*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)*cos(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-
b)/(a+b))^(1/2))+2*C*cos(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))*a^3)/(b+a*cos(d*x+c))/sin(d*x+c)/a^2/(a
+b)/(a-b)
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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(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(3/2),x)
[Out]
int((cos(c + d*x)*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(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)*cos(c + d*x)/(a + b*sec(c + d*x))**(3/2), x)
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