3.864 \(\int \frac {(a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=121 \[ \frac {2 \left (3 a^2 C+6 a b B+b^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (a^2 B-2 a b C-b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 b^2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

[Out]

-2*(B*a^2-B*b^2-2*C*a*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))
/d+2/3*(6*B*a*b+3*C*a^2+C*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))/d+2*a^2*B*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*b^2*C*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.34, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3029, 2988, 3023, 2748, 2641, 2639} \[ \frac {2 \left (3 a^2 C+6 a b B+b^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (a^2 B-2 a b C-b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 b^2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a^2*B - b^2*B - 2*a*b*C)*EllipticE[(c + d*x)/2, 2])/d + (2*(6*a*b*B + 3*a^2*C + b^2*C)*EllipticF[(c + d*x
)/2, 2])/(3*d) + (2*a^2*B*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*b^2*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*
d)

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 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 2988

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -1]

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]

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\int \frac {(a+b \cos (c+d x))^2 (B+C \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-2 \int \frac {-\frac {1}{2} a (2 b B+a C)+\frac {1}{2} \left (a^2 B-b^2 B-2 a b C\right ) \cos (c+d x)-\frac {1}{2} b^2 C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 b^2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {4}{3} \int \frac {\frac {1}{4} \left (-b^2 C-3 a (2 b B+a C)\right )+\frac {3}{4} \left (a^2 B-b^2 B-2 a b C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 b^2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\left (a^2 B-b^2 B-2 a b C\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{3} \left (-6 a b B-3 a^2 C-b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 \left (a^2 B-b^2 B-2 a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (6 a b B+3 a^2 C+b^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 b^2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 102, normalized size = 0.84 \[ \frac {2 \left (\left (3 a^2 C+6 a b B+b^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (-3 a^2 B+6 a b C+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {\sin (c+d x) \left (3 a^2 B+b^2 C \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-3*a^2*B + 3*b^2*B + 6*a*b*C)*EllipticE[(c + d*x)/2, 2] + (6*a*b*B + 3*a^2*C + b^2*C)*EllipticF[(c + d*x)
/2, 2] + ((3*a^2*B + b^2*C*Cos[c + d*x])*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(3*d)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{2} \cos \left (d x + c\right )^{3} + B a^{2} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*cos(d*x + c)^3 + B*a^2 + (2*C*a*b + B*b^2)*cos(d*x + c)^2 + (C*a^2 + 2*B*a*b)*cos(d*x + c))/co
s(d*x + c)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.30, size = 404, normalized size = 3.34 \[ -\frac {2 \left (4 C \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 B a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-6 B \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+b^{2} C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -2 C \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(4*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*B*a*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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))*a^2-3*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))*b^2-6*B*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+3*a^2*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*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))-6*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-2*C*b^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*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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.82, size = 158, normalized size = 1.31 \[ \frac {C\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*b^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*B*b^2*ellipticE(c/2
+ (d*x)/2, 2))/d + (2*C*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (4*B*a*b*ellipticF(c/2 + (d*x)/2, 2))/d + (4*C*a
*b*ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(
c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))

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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))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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