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

Optimal. Leaf size=126 \[ \frac {(5 B+19 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(5 B-13 C) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+1/16*(5*B-13*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/32*(5*B
+19*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3019, 2750, 2649, 206} \[ \frac {(5 B+19 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(5 B-13 C) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(B-C) \sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*B + 19*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((B
- C)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((5*B - 13*C)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x]
)^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

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

Rule 2750

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

Rule 3019

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 - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(B-C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {\int \frac {-\frac {5}{2} a (B-C)-4 a C \cos (c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(B-C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 B-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(5 B+19 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(B-C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 B-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(5 B+19 C) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {(5 B+19 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(B-C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(5 B-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 87, normalized size = 0.69 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) ((5 B-13 C) \cos (c+d x)+B-9 C)+2 (5 B+19 C) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{16 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*B + 19*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (B - 9*C + (5*B - 13*C)*Cos[c + d*x])*Tan[(c +
d*x)/2])/(16*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

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fricas [B]  time = 0.50, size = 223, normalized size = 1.77 \[ \frac {\sqrt {2} {\left ({\left (5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, B + 19 \, C\right )} \cos \left (d x + c\right ) + 5 \, B + 19 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (5 \, B - 13 \, C\right )} \cos \left (d x + c\right ) + B - 9 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(2)*((5*B + 19*C)*cos(d*x + c)^3 + 3*(5*B + 19*C)*cos(d*x + c)^2 + 3*(5*B + 19*C)*cos(d*x + c) + 5*B
+ 19*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*
x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((5*B - 13*C)*cos(d*x + c) + B - 9*C)*sqrt(a*cos(d*x
+ c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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giac [A]  time = 1.40, size = 134, normalized size = 1.06 \[ -\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (B a^{5} - C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} - \frac {\sqrt {2} {\left (3 \, B a^{5} - 11 \, C a^{5}\right )}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {\sqrt {2} {\left (5 \, B + 19 \, C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(B*a^5 - C*a^5)*tan(1/2*d*x + 1/2*c)^2/a^8 - sqrt(2)*(3*B
*a^5 - 11*C*a^5)/a^8)*tan(1/2*d*x + 1/2*c) + sqrt(2)*(5*B + 19*C)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt
(a*tan(1/2*d*x + 1/2*c)^2 + a)))/a^(5/2))/d

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maple [B]  time = 1.34, size = 292, normalized size = 2.32 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (5 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +19 C \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +5 B \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+2 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2
*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a+19*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(
1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a+5*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/
2*d*x+1/2*c)^2-13*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2-2*B*2^(1/2)*(a*sin(1/2
*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/sin(1/2*d*x+1/2*c)/(a
*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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