3.211 \(\int \frac{\sqrt{\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=201 \[ -\frac{(5 A-17 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(13 A+7 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A+3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{5/2}}+\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}} \]

[Out]

((13*A + 7*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2
]*a^(7/2)*d) + ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((A + 3*B)*Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)) - ((5*A - 17*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.579293, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2977, 2978, 12, 2782, 205} \[ -\frac{(5 A-17 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{192 a^2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(13 A+7 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A+3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \cos (c+d x)+a)^{5/2}}+\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((13*A + 7*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2
]*a^(7/2)*d) + ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)) + ((A + 3*B)*Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)) - ((5*A - 17*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2))

Rule 2977

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

Rule 2978

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx &=\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{\int \frac{\frac{1}{2} a (A-B)+2 a (A+2 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{4} a^2 (11 A+B)+\frac{3}{2} a^2 (A+3 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-17 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{3 a^3 (13 A+7 B)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-17 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(13 A+7 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-17 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}-\frac{(13 A+7 B) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{64 a^2 d}\\ &=\frac{(13 A+7 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{(A+3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-17 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.99237, size = 215, normalized size = 1.07 \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{8} \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) (4 (A+35 B) \cos (c+d x)+(17 B-5 A) \cos (2 (c+d x))+73 A+59 B)+\frac{3 i (13 A+7 B) e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{24 d (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^7*(((3*I)*(13*A + 7*B)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*A
rcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[1 + E^((2*I)*(c + d*x))] + (Sqrt[C
os[c + d*x]]*(73*A + 59*B + 4*(A + 35*B)*Cos[c + d*x] + (-5*A + 17*B)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^5*Tan
[(c + d*x)/2])/8))/(24*d*(a*(1 + Cos[c + d*x]))^(7/2))

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Maple [B]  time = 0.62, size = 549, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(7/2),x)

[Out]

1/384/d*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^4*(10*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*co
s(d*x+c)^4-39*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)-4*A*(cos(d*x+c)/(1+cos(d*x+
c)))^(3/2)*cos(d*x+c)^3-21*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)-78*A*arcsin((-
1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-88*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2-
34*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4-42*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*cos(d*x+c)
^2*sin(d*x+c)-39*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*2^(1/2)*cos(d*x+c)+4*A*cos(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(3/2)-106*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3-21*B*arcsin((-1+cos(d*x+c))/sin(d*
x+c))*sin(d*x+c)*2^(1/2)*cos(d*x+c)+78*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+98*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*cos(d*x+c)^2+42*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c))/sin(d*x+c)^9/(cos(d*x+c)/(1+cos(d*x+c)))^(
3/2)/a^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65976, size = 705, normalized size = 3.51 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 13 \, A + 7 \, B\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (5 \, A - 17 \, B\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (A + 35 \, B\right )} \cos \left (d x + c\right ) - 39 \, A - 21 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/384*(3*sqrt(2)*((13*A + 7*B)*cos(d*x + c)^4 + 4*(13*A + 7*B)*cos(d*x + c)^3 + 6*(13*A + 7*B)*cos(d*x + c)^2
+ 4*(13*A + 7*B)*cos(d*x + c) + 13*A + 7*B)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(c
os(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((5*A - 17*B)*cos(d*x + c)^2 - 2*(A + 35*B)
*cos(d*x + c) - 39*A - 21*B)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 +
4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*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)**(1/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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