3.256 \(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=183 \[ \frac {2 b^3 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 (7 A+5 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 d}+\frac {6 b^2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b B \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d} \]

[Out]

2/5*b*B*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*C*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/21*b^3*(7*A+5*C)*(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))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))
^(1/2)+2/21*b^2*(7*A+5*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+6/5*b^2*B*(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*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {16, 3023, 2748, 2635, 2642, 2641, 2640, 2639} \[ \frac {2 b^2 (7 A+5 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 d}+\frac {2 b^3 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {6 b^2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b B \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x],x]

[Out]

(6*b^2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b^3*(7*A + 5*C)*Sqrt[Co
s[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (2*b^2*(7*A + 5*C)*Sqrt[b*Cos[c + d*x]]*S
in[c + d*x])/(21*d) + (2*b*B*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + (2*C*(b*Cos[c + d*x])^(5/2)*Sin[c +
d*x])/(7*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=b \int (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {2}{7} \int (b \cos (c+d x))^{3/2} \left (\frac {1}{2} b (7 A+5 C)+\frac {7}{2} b B \cos (c+d x)\right ) \, dx\\ &=\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+B \int (b \cos (c+d x))^{5/2} \, dx+\frac {1}{7} (b (7 A+5 C)) \int (b \cos (c+d x))^{3/2} \, dx\\ &=\frac {2 b^2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{5} \left (3 b^2 B\right ) \int \sqrt {b \cos (c+d x)} \, dx+\frac {1}{21} \left (b^3 (7 A+5 C)\right ) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx\\ &=\frac {2 b^2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {\left (b^3 (7 A+5 C) \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}}+\frac {\left (3 b^2 B \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=\frac {6 b^2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^3 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 b^2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 109, normalized size = 0.60 \[ \frac {b (b \cos (c+d x))^{3/2} \left (\sin (c+d x) \sqrt {\cos (c+d x)} (70 A+42 B \cos (c+d x)+15 C \cos (2 (c+d x))+65 C)+10 (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+126 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{105 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x],x]

[Out]

(b*(b*Cos[c + d*x])^(3/2)*(126*B*EllipticE[(c + d*x)/2, 2] + 10*(7*A + 5*C)*EllipticF[(c + d*x)/2, 2] + Sqrt[C
os[c + d*x]]*(70*A + 65*C + 42*B*Cos[c + d*x] + 15*C*Cos[2*(c + d*x)])*Sin[c + d*x]))/(105*d*Cos[c + d*x]^(3/2
))

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + B b^{2} \cos \left (d x + c\right )^{3} + A b^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4 + B*b^2*cos(d*x + c)^3 + A*b^2*cos(d*x + c)^2)*sqrt(b*cos(d*x + c))*sec(d*x + c
), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c), x)

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maple [A]  time = 1.52, size = 353, normalized size = 1.93 \[ -\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (240 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-168 B -360 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (140 A +168 B +280 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-70 A -42 B -80 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x)

[Out]

-2/105*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(240*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^8+(-168*B-360*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A+168*B+280*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*
d*x+1/2*c)+(-70*A-42*B-80*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-63*B*(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))+25*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(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)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x),x)

[Out]

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x), 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((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),x)

[Out]

Timed out

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