3.601 \(\int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx\)
Optimal. Leaf size=328 \[ \frac {b \left (11 a^2-5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {b \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}} \]
[Out]
-1/4*(8*a^4-29*a^2*b^2+15*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^
(1/2))/a^3/(a^2-b^2)^2/d+1/4*b*(11*a^2-5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/
2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2)^2/d-1/4*b*(35*a^4-38*a^2*b^2+15*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*
d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a^3/(a-b)^2/(a+b)^3/d+1/4*(8*a^4-29*a^2*b^2+15*b^4
)*sin(d*x+c)/a^3/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)+1/2*b^2*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2/cos(d*x+c)
^(1/2)+1/4*b^2*(11*a^2-5*b^2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))/cos(d*x+c)^(1/2)
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Rubi [A] time = 1.07, antiderivative size = 328, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used =
{2802, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {b \left (11 a^2-5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac {\left (-29 a^2 b^2+8 a^4+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {b \left (-38 a^2 b^2+35 a^4+15 b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d (a-b)^2 (a+b)^3}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\left (-29 a^2 b^2+8 a^4+15 b^4\right ) \sin (c+d x)}{4 a^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^3),x]
[Out]
-((8*a^4 - 29*a^2*b^2 + 15*b^4)*EllipticE[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d) + (b*(11*a^2 - 5*b^2)*Ellip
ticF[(c + d*x)/2, 2])/(4*a^2*(a^2 - b^2)^2*d) - (b*(35*a^4 - 38*a^2*b^2 + 15*b^4)*EllipticPi[(2*b)/(a + b), (c
+ d*x)/2, 2])/(4*a^3*(a - b)^2*(a + b)^3*d) + ((8*a^4 - 29*a^2*b^2 + 15*b^4)*Sin[c + d*x])/(4*a^3*(a^2 - b^2)
^2*d*Sqrt[Cos[c + d*x]]) + (b^2*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2) +
(b^2*(11*a^2 - 5*b^2)*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x]))
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2802
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 3059
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps
\begin {align*} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (4 a^2-5 b^2\right )-2 a b \cos (c+d x)+\frac {3}{2} b^2 \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (8 a^4-29 a^2 b^2+15 b^4\right )-a b \left (4 a^2-b^2\right ) \cos (c+d x)+\frac {1}{4} b^2 \left (11 a^2-5 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\int \frac {-\frac {3}{8} b \left (8 a^4-11 a^2 b^2+5 b^4\right )-\frac {1}{2} a \left (2 a^4-10 a^2 b^2+5 b^4\right ) \cos (c+d x)-\frac {1}{8} b \left (8 a^4-29 a^2 b^2+15 b^4\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}-\frac {\int \frac {\frac {3}{8} b^2 \left (8 a^4-11 a^2 b^2+5 b^4\right )-\frac {1}{8} a b^3 \left (11 a^2-5 b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3 b \left (a^2-b^2\right )^2}-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\left (b \left (11 a^2-5 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}-\frac {\left (b \left (35 a^4-38 a^2 b^2+15 b^4\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {b \left (11 a^2-5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {b \left (35 a^4-38 a^2 b^2+15 b^4\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x)}{4 a^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 3.62, size = 334, normalized size = 1.02 \[ \frac {4 \sqrt {\cos (c+d x)} \left (\frac {b^3 \sin (c+d x) \left (-15 a^3+\left (7 b^3-13 a^2 b\right ) \cos (c+d x)+9 a b^2\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+8 \tan (c+d x)\right )-\frac {\frac {2 \left (56 a^4 b-95 a^2 b^3+45 b^5\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 a \left (2 a^4-10 a^2 b^2+5 b^4\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}+\frac {2 \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 a^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^3),x]
[Out]
(-(((2*(56*a^4*b - 95*a^2*b^3 + 45*b^5)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*a*(2*a^4 - 10*
a^2*b^2 + 5*b^4)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b +
(2*(8*a^4 - 29*a^2*b^2 + 15*b^4)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[Arc
Sin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x]
)/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2)) + 4*Sqrt[Cos[c + d*x]]*((b^3*(-15*a^3 + 9*a*b^2 + (-13*a^
2*b + 7*b^3)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + 8*Tan[c + d*x]))/(16*a^3*d)
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fricas [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(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
integrate(1/((b*cos(d*x + c) + a)^3*cos(d*x + c)^(3/2)), x)
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maple [B] time = 3.71, size = 1992, normalized size = 6.07 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/a*b*(-1/2*b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a
^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-
b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))*b+3/8/(a+b)/(a^2-b^2)/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2
)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(a^2-b^
2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(
1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^
(1/2))+9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(c
os(1/2*d*x+1/2*c),2^(1/2))-15/4*a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+
1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b
),2^(1/2))+3/2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-3/4/a^2/
(a^2-b^2)^2/(-2*a*b+2*b^2)*b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))+4*b^2/a^3/(-2*a*b+2*b^
2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^
2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2/a^3*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+
2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/
2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)-2/a^2*b*(-b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/
2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+
1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1
/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2
*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b
^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4
+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2
*d*x+1/2*c)^2-1)^(1/2)/d
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^3),x)
[Out]
int(1/(cos(c + d*x)^(3/2)*(a + b*cos(c + d*x))^3), x)
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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(1/cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**3,x)
[Out]
Timed out
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