3.507 \(\int (a+b \cos (c+d x))^{5/2} \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=323 \[ \frac {\left (16 a^2+33 b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{24 d}+\frac {a \left (16 a^2+59 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {5 b \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}+\frac {13 a b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{12 d} \]

[Out]

-1/24*(16*a^2+33*b^2)*(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)*(b/
(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/24*a*(16*a^2+59*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)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^
(1/2)/d/(a+b*cos(d*x+c))^(1/2)+5/8*b*(4*a^2+b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(si
n(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)+1/24*(16*a
^2+33*b^2)*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d+13/12*a*b*sec(d*x+c)*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d+1/3*a^
2*sec(d*x+c)^2*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 1.17, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2792, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac {\left (16 a^2+33 b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{24 d}+\frac {a \left (16 a^2+59 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \cos (c+d x)}}-\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {5 b \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}+\frac {13 a b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^4,x]

[Out]

-((16*a^2 + 33*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(24*d*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]) + (a*(16*a^2 + 59*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)
])/(24*d*Sqrt[a + b*Cos[c + d*x]]) + (5*b*(4*a^2 + b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c +
d*x)/2, (2*b)/(a + b)])/(8*d*Sqrt[a + b*Cos[c + d*x]]) + ((16*a^2 + 33*b^2)*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d
*x])/(24*d) + (13*a*b*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(12*d) + (a^2*Sqrt[a + b*Cos[c + d*x
]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 2653

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

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], 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[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[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 (a+b \cos (c+d x))^{5/2} \sec ^4(c+d x) \, dx &=\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \frac {\left (\frac {13 a^2 b}{2}+a \left (2 a^2+9 b^2\right ) \cos (c+d x)+\frac {3}{2} b \left (a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {13 a b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \frac {\left (\frac {1}{4} a^2 \left (16 a^2+33 b^2\right )+\frac {1}{2} a b \left (19 a^2+12 b^2\right ) \cos (c+d x)+\frac {13}{4} a^2 b^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a}\\ &=\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {13 a b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \frac {\left (\frac {15}{8} a^2 b \left (4 a^2+b^2\right )+\frac {13}{4} a^3 b^2 \cos (c+d x)-\frac {1}{8} a^2 b \left (16 a^2+33 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {13 a b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\int \frac {\left (-\frac {15}{8} a^2 b^2 \left (4 a^2+b^2\right )-\frac {1}{8} a^3 b \left (16 a^2+59 b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a^2 b}+\frac {1}{48} \left (-16 a^2-33 b^2\right ) \int \sqrt {a+b \cos (c+d x)} \, dx\\ &=\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {13 a b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{16} \left (5 b \left (4 a^2+b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{48} \left (a \left (16 a^2+59 b^2\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx+\frac {\left (\left (-16 a^2-33 b^2\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{48 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=-\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {13 a b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\left (5 b \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{16 \sqrt {a+b \cos (c+d x)}}+\frac {\left (a \left (16 a^2+59 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{48 \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a \left (16 a^2+59 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \cos (c+d x)}}+\frac {5 b \left (4 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (16 a^2+33 b^2\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {13 a b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {a^2 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C]  time = 4.12, size = 434, normalized size = 1.34 \[ \frac {\frac {2 b \left (104 a^2-3 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+4 \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)} \left (\left (8 a^2+\frac {33 b^2}{2}\right ) \sin (2 (c+d x))+8 a^2 \tan (c+d x)+26 a b \sin (c+d x)\right )-\frac {2 i \left (16 a^2+33 b^2\right ) \csc (c+d x) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\cos (c+d x)+1)}{a-b}} \left (b \left (b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )-2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+\frac {104 a b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^4,x]

[Out]

((104*a*b^2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]]
+ (2*b*(104*a^2 - 3*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a
+ b*Cos[c + d*x]] - ((2*I)*(16*a^2 + 33*b^2)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c +
d*x]))/(a - b))]*Csc[c + d*x]*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]],
(a + b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)
] + b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b*
Sqrt[-(a + b)^(-1)]) + 4*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*(26*a*b*Sin[c + d*x] + (8*a^2 + (33*b^2)/2)*S
in[2*(c + d*x)] + 8*a^2*Tan[c + d*x]))/(96*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((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)

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maple [B]  time = 1.11, size = 1742, normalized size = 5.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x)

[Out]

-1/24*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((256*a^2*b+528*b^3)*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^8+(-128*a^3-384*a^2*b-472*a*b^2-792*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(128*a^3+328*a
^2*b+472*a*b^2+396*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-48*a^3-100*a^2*b-118*a*b^2-66*b^3)*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-8*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(16*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+59*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2
))*a*b^2-16*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+16*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^
(1/2))*a^2*b-33*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+33*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))*b^3-60*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2*b-15*EllipticPi(cos(1/2*d*x+1/2*c)
,2,(-2*b/(a-b))^(1/2))*b^3)*sin(1/2*d*x+1/2*c)^6+12*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(16*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+59*EllipticF(cos(1/2*d*x+1/2*c
),(-2*b/(a-b))^(1/2))*a*b^2-16*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+16*EllipticE(cos(1/2*d*x+1
/2*c),(-2*b/(a-b))^(1/2))*a^2*b-33*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+33*EllipticE(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-60*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2*b-15*EllipticPi
(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^3)*sin(1/2*d*x+1/2*c)^4-6*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(
a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+59*EllipticF
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-16*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+16*Ellip
ticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-33*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2+33
*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-60*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a
^2*b-15*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^3)*sin(1/2*d*x+1/2*c)^2+16*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
^3+59*b^2*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))-16*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^
(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+16*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2
*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b-33*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))
*a*b^2+33*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d
*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-60*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)
/(a-b))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-15*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a
-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))/(2*cos(1/2*d*
x+1/2*c)^2-1)^3/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*
x+1/2*c)^2*b+a+b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^4,x)

[Out]

int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^4, 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((a+b*cos(d*x+c))**(5/2)*sec(d*x+c)**4,x)

[Out]

Timed out

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