3.113 \(\int \frac{\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=169 \[ \frac{(3 A+7 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{6 a^2 d}+\frac{(3 A+11 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(3 A+13 C) \sin (c+d x)}{3 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

((3*A + 11*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A
+ C)*Cos[c + d*x]^2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((3*A + 13*C)*Sin[c + d*x])/(3*a*d*Sqrt[a
+ a*Cos[c + d*x]]) + ((3*A + 7*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*a^2*d)

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Rubi [A]  time = 0.341496, antiderivative size = 169, 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, 3023, 2751, 2649, 206} \[ \frac{(3 A+7 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{6 a^2 d}+\frac{(3 A+11 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(3 A+13 C) \sin (c+d x)}{3 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

((3*A + 11*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A
+ C)*Cos[c + d*x]^2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((3*A + 13*C)*Sin[c + d*x])/(3*a*d*Sqrt[a
+ a*Cos[c + d*x]]) + ((3*A + 7*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*a^2*d)

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 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 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))^{3/2}} \, dx &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\cos (c+d x) \left (-2 a C+\frac{1}{2} a (3 A+7 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{-2 a C \cos (c+d x)+\frac{1}{2} a (3 A+7 C) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(3 A+7 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac{\int \frac{\frac{1}{4} a^2 (3 A+7 C)-\frac{1}{2} a^2 (3 A+13 C) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(3 A+13 C) \sin (c+d x)}{3 a d \sqrt{a+a \cos (c+d x)}}+\frac{(3 A+7 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac{(3 A+11 C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(3 A+13 C) \sin (c+d x)}{3 a d \sqrt{a+a \cos (c+d x)}}+\frac{(3 A+7 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}-\frac{(3 A+11 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 )}{2 a d}\\ &=\frac{(3 A+11 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(3 A+13 C) \sin (c+d x)}{3 a d \sqrt{a+a \cos (c+d x)}}+\frac{(3 A+7 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.605505, size = 94, normalized size = 0.56 \[ \frac{3 (3 A+11 C) \cos \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\tan \left (\frac{1}{2} (c+d x)\right ) (3 A+12 C \cos (c+d x)-2 C \cos (2 (c+d x))+17 C)}{6 a d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(3*(3*A + 11*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] - (3*A + 17*C + 12*C*Cos[c + d*x] - 2*C*Cos[2*(c +
d*x)])*Tan[(c + d*x)/2])/(6*a*d*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [A]  time = 0.065, size = 292, normalized size = 1.7 \begin{align*}{\frac{1}{12\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 16\,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}+9\,A\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 ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+33\,C\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 ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-40\,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}-3\,A\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a}-3\,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 ) ^{-1}{a}^{-{\frac{5}{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))^(3/2),x)

[Out]

1/12*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*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+
9*A*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)^2*a+33*
C*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)^2*a-40*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)^2-3*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)
*a^(1/2)-3*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)/a^(5/2)/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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.68336, size = 533, normalized size = 3.15 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 11 \, 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 (4 \, C \cos \left (d x + c\right )^{2} - 12 \, C \cos \left (d x + c\right ) - 3 \, A - 19 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{24 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

1/24*(3*sqrt(2)*((3*A + 11*C)*cos(d*x + c)^2 + 2*(3*A + 11*C)*cos(d*x + c) + 3*A + 11*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*(4*C*cos(d*x + c)^2 - 12*C*cos(d*x + c) - 3*A - 19*C)*sqrt(a*cos(d*x + c) + a)*sin(
d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.9682, size = 225, normalized size = 1.33 \begin{align*} -\frac{\frac{3 \,{\left (3 \, \sqrt{2} A + 11 \, \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{3}{2}}} + \frac{{\left ({\left (\frac{3 \,{\left (\sqrt{2} A a + \sqrt{2} C a\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a} + \frac{2 \,{\left (3 \, \sqrt{2} A a + 23 \, \sqrt{2} C a\right )}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{3 \,{\left (\sqrt{2} A a + 9 \, \sqrt{2} C a\right )}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}}}}{12 \, 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))^(3/2),x, algorithm="giac")

[Out]

-1/12*(3*(3*sqrt(2)*A + 11*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^(3/2) + ((3*(sqrt(2)*A*a + sqrt(2)*C*a)*tan(1/2*d*x + 1/2*c)^2/a + 2*(3*sqrt(2)*A*a + 23*sqrt(2)*C*a)/a
)*tan(1/2*d*x + 1/2*c)^2 + 3*(sqrt(2)*A*a + 9*sqrt(2)*C*a)/a)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 +
a)^(3/2))/d