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

Optimal. Leaf size=165 \[ \frac{(A+9 C) \sin (c+d x)}{4 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{5 (A-15 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac{(3 A-13 C) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(5*(A - 15*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A
+ C)*Cos[c + d*x]^2*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((3*A - 13*C)*Sin[c + d*x])/(16*a*d*(a +
a*Cos[c + d*x])^(3/2)) + ((A + 9*C)*Sin[c + d*x])/(4*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.367333, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3042, 2968, 3019, 2751, 2649, 206} \[ \frac{(A+9 C) \sin (c+d x)}{4 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{5 (A-15 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac{(3 A-13 C) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(A - 15*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A
+ C)*Cos[c + d*x]^2*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((3*A - 13*C)*Sin[c + d*x])/(16*a*d*(a +
a*Cos[c + d*x])^(3/2)) + ((A + 9*C)*Sin[c + d*x])/(4*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 3042

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

Rule 2968

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

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]

Rule 2751

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

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 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])

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\cos (c+d x) \left (2 a (A-C)+\frac{1}{2} a (A+9 C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{2 a (A-C) \cos (c+d x)+\frac{1}{2} a (A+9 C) \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(3 A-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{\int \frac{-\frac{3}{4} a^2 (3 A-13 C)-a^2 (A+9 C) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(3 A-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(A+9 C) \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(5 (A-15 C)) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(3 A-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(A+9 C) \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(5 (A-15 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 (A-15 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{(A+C) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(3 A-13 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(A+9 C) \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.699797, size = 95, normalized size = 0.58 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) (5 (A+17 C) \cos (c+d x)+A+16 C \cos (2 (c+d x))+65 C)+10 (A-15 C) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{16 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(A - 15*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (A + 65*C + 5*(A + 17*C)*Cos[c + d*x] + 16*C*Cos
[2*(c + d*x)])*Tan[(c + d*x)/2])/(16*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 0.071, size = 327, normalized size = 2. \begin{align*}{\frac{1}{32\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 5\,A\sqrt{2}\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-75\,C\sqrt{2}\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+64\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+5\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+21\,C\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-2\,A\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a}-2\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{a}^{-{\frac{7}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{a \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+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*A*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-75*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+64*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)^4+5*A*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+21*C*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)^2-2*A*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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 1.91618, size = 610, normalized size = 3.7 \begin{align*} -\frac{5 \, \sqrt{2}{\left ({\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (A - 15 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (A - 15 \, C\right )} \cos \left (d x + c\right ) + A - 15 \, 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 (32 \, C \cos \left (d x + c\right )^{2} + 5 \,{\left (A + 17 \, C\right )} \cos \left (d x + c\right ) + A + 49 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(5*sqrt(2)*((A - 15*C)*cos(d*x + c)^3 + 3*(A - 15*C)*cos(d*x + c)^2 + 3*(A - 15*C)*cos(d*x + c) + A - 15
*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*(32*C*cos(d*x + c)^2 + 5*(A + 17*C)*cos(d*x + c) + A + 49*
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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 2.1808, size = 240, normalized size = 1.45 \begin{align*} -\frac{\frac{{\left ({\left (\frac{2 \,{\left (\sqrt{2} A a^{6} + \sqrt{2} C a^{6}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8}} - \frac{\sqrt{2} A a^{6} + 17 \, \sqrt{2} C a^{6}}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{3 \, \sqrt{2} A a^{6} + 83 \, \sqrt{2} C a^{6}}{a^{8}}\right )} \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}} + \frac{5 \,{\left (\sqrt{2} A - 15 \, \sqrt{2} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(((2*(sqrt(2)*A*a^6 + sqrt(2)*C*a^6)*tan(1/2*d*x + 1/2*c)^2/a^8 - (sqrt(2)*A*a^6 + 17*sqrt(2)*C*a^6)/a^8
)*tan(1/2*d*x + 1/2*c)^2 - (3*sqrt(2)*A*a^6 + 83*sqrt(2)*C*a^6)/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x +
1/2*c)^2 + a) + 5*(sqrt(2)*A - 15*sqrt(2)*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