3.430 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=123 \[ -\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-2/5*(3*A+5*C)*(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*B*(
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+2/5*A*sin(d*x+c)/d/cos(
d*x+c)^(5/2)+2/3*B*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(3*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3021, 2748, 2636, 2641, 2639} \[ -\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (3 A+5 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Cos[c + d*x]^(7/2),x]

[Out]

(-2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*A*Sin[c + d*x])/
(5*d*Cos[c + d*x]^(5/2)) + (2*B*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(3*A + 5*C)*Sin[c + d*x])/(5*d*Sqr
t[Cos[c + d*x]])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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 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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 112, normalized size = 0.91 \[ \frac {-6 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+9 A \sin (2 (c+d x))+6 A \tan (c+d x)+10 B \sin (c+d x)+10 B \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+15 C \sin (2 (c+d x))}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Cos[c + d*x]^(7/2),x]

[Out]

(-6*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2,
2] + 10*B*Sin[c + d*x] + 9*A*Sin[2*(c + d*x)] + 15*C*Sin[2*(c + d*x)] + 6*A*Tan[c + d*x])/(15*d*Cos[c + d*x]^(
3/2))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/cos(d*x + c)^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/cos(d*x + c)^(7/2), x)

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maple [B]  time = 4.91, size = 799, normalized size = 6.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)

[Out]

2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4
+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^3*(36*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
20*B*EllipticF(cos(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)*sin(1
/2*d*x+1/2*c)^4+60*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))*sin(1/2*d*x+1/2*c)^4-120*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*A*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+72*A*cos(1/2*d*x
+1/2*c)*sin(1/2*d*x+1/2*c)^4-20*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-60*C*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+120*
C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+9*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)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))-24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+5*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))-10*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^2+15*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))-30*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)*(-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)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/cos(d*x + c)^(7/2), x)

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mupad [B]  time = 2.11, size = 108, normalized size = 0.88 \[ \frac {6\,A\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+10\,B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,C\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/cos(c + d*x)^(7/2),x)

[Out]

(6*A*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 10*B*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/
4, 1/2], 1/4, cos(c + d*x)^2) + 30*C*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/
(15*d*cos(c + d*x)^(5/2)*(1 - cos(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)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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