3.729 \(\int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=478 \[ \frac {(a-b) \sqrt {a+b} \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}+\frac {\sqrt {a+b} \left (30 a^3 C+a^2 b (15 A-46 C)+2 a b^2 (45 A+17 C)-6 b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}-\frac {b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {a b (15 A-16 C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}-\frac {5 a A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d} \]
[Out]
A*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/15*(a-b)*(a^2*(15*A-46*C)-6*b^2*(5*A+3*C))*cot(d*x+c)*EllipticE((a+b*s
ec(d*x+c))^(1/2)/(a+b)^(1/2),((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)/b/d+1/15*(a^2*b*(15*A-46*C)+30*a^3*C-6*b^3*(5*A+3*C)+2*a*b^2*(45*A+17*C))*cot(d*x+c)*EllipticF(
(a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((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)/b/d-5*a*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)/d-1/5*b*(5*A-2*C)*(a+b*sec
(d*x+c))^(3/2)*tan(d*x+c)/d-1/15*a*b*(15*A-16*C)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A] time = 0.81, antiderivative size = 478, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used =
{4095, 4056, 4058, 3921, 3784, 3832, 4004} \[ \frac {\sqrt {a+b} \left (a^2 b (15 A-46 C)+30 a^3 C+2 a b^2 (45 A+17 C)-6 b^3 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}+\frac {(a-b) \sqrt {a+b} \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}-\frac {b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {a b (15 A-16 C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d}-\frac {5 a A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
((a - b)*Sqrt[a + b]*(a^2*(15*A - 46*C) - 6*b^2*(5*A + 3*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
))])/(15*b*d) + (Sqrt[a + b]*(a^2*b*(15*A - 46*C) + 30*a^3*C - 6*b^3*(5*A + 3*C) + 2*a*b^2*(45*A + 17*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))])/(15*b*d) - (5*a*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)]*Sqr
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/d + (A*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/d - (a*b*(15*A - 16*C)*S
qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) - (b*(5*A - 2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
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 4056
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int
[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) + a
*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 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 4095
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*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e +
f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]
Rubi steps
\begin {align*} \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^{3/2} \left (\frac {5 A b}{2}+a C \sec (c+d x)-\frac {1}{2} b (5 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {a+b \sec (c+d x)} \left (\frac {25 a A b}{4}+\frac {1}{2} \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \sec (c+d x)-\frac {1}{4} a b (15 A-16 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a b (15 A-16 C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {75}{8} a^2 A b+\frac {1}{4} a \left (45 A b^2+15 a^2 C+17 b^2 C\right ) \sec (c+d x)-\frac {1}{8} b \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a b (15 A-16 C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {75}{8} a^2 A b+\left (\frac {1}{4} a \left (45 A b^2+15 a^2 C+17 b^2 C\right )+\frac {1}{8} b \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx-\frac {1}{30} \left (b \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {(a-b) \sqrt {a+b} \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 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}}}{15 b d}+\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a b (15 A-16 C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{2} \left (5 a^2 A b\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{30} \left (a^2 b (15 A-46 C)+30 a^3 C-6 b^3 (5 A+3 C)+2 a b^2 (45 A+17 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {(a-b) \sqrt {a+b} \left (a^2 (15 A-46 C)-6 b^2 (5 A+3 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}}}{15 b d}+\frac {\sqrt {a+b} \left (a^2 b (15 A-46 C)+30 a^3 C-6 b^3 (5 A+3 C)+2 a b^2 (45 A+17 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}}}{15 b d}-\frac {5 a 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}}}{d}+\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a b (15 A-16 C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}
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Mathematica [B] time = 26.42, size = 6785, normalized size = 14.19 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]
[Out]
Result too large to show
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fricas [F] time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A a^{2} \cos \left (d x + c\right ) + {\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*b^2*cos(d*x + c)*sec(d*x + c)^4 + 2*C*a*b*cos(d*x + c)*sec(d*x + c)^3 + 2*A*a*b*cos(d*x + c)*sec(d
*x + c) + A*a^2*cos(d*x + c) + (C*a^2 + A*b^2)*cos(d*x + c)*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c), x)
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maple [B] time = 2.44, size = 3498, normalized size = 7.32 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)
[Out]
-1/15/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(10*C*cos(d*x+c)^3*a*b^2+30*A*c
os(d*x+c)^3*b^3-6*b^3*C+46*C*cos(d*x+c)^4*a^3-30*A*sin(d*x+c)*cos(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))*a*b^2-3
0*A*cos(d*x+c)^2*b^3-30*A*cos(d*x+c)^3*a*b^2+15*A*cos(d*x+c)^4*a^2*b+30*A*cos(d*x+c)^4*a*b^2-15*A*cos(d*x+c)^3
*a^2*b-68*C*cos(d*x+c)^2*a^2*b-28*C*cos(d*x+c)*a*b^2+22*C*cos(d*x+c)^4*a^2*b+18*C*cos(d*x+c)^4*a*b^2+46*C*cos(
d*x+c)^3*a^2*b-46*C*cos(d*x+c)^3*a^3+18*C*cos(d*x+c)^3*b^3-12*C*cos(d*x+c)^2*b^3+15*A*cos(d*x+c)^5*a^3+15*A*co
s(d*x+c)^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)*Elliptic
E((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+15*A*cos(d*x+c)^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
))*a^2*b-90*A*cos(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)*Ell
ipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b+150*A*cos(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)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(
a+b))^(1/2))*sin(d*x+c)*a^2*b-90*A*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))*sin(d*x+c)*a^2*b+150*A*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)*EllipticPi((-1+cos(d*x+c))/s
in(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b+90*A*sin(d*x+c)*cos(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*
b^2-46*C*sin(d*x+c)*cos(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))*a^2*b-18*C*sin(d*x+c)*cos(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))*a*b^2+46*C*sin(d*x+c)*cos(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+34*C*sin(d*x+c)*cos(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)*EllipticF((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-30*A*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^2+90*A*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^2-46*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a
^2*b-18*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+46*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^2*b+34*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*b^2-30*A*sin(d*x+c)*cos(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))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*b^3+30*A*sin(d*x+c)*cos(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-15*A*cos(d*x+c
)^4*a^3-46*C*sin(d*x+c)*cos(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))*a^3-18*C*sin(d*x+c)*cos(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+18*C*sin(d*x+c)*cos(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-30*A*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))*b^3+30*A*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))*b^3-46*C*sin(d*x+c)*c
os(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^3-18*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+18*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)*Ellipt
icF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+15*A*cos(d*x+c)^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+30*C*cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3+15*A*cos(d*x+c)^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))*a^3+30*C*cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3)/(b+a*cos(d*x+c))/cos(d*x+c)^2
/sin(d*x+c)^5
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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+b*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \cos \left (c+d\,x\right )\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]
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))^(5/2),x)
[Out]
int(cos(c + d*x)*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(5/2), x)
________________________________________________________________________________________
sympy [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+b*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)
[Out]
Timed out
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