3.2.100 \(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\) [200]

Optimal. Leaf size=156 \[ -\frac {(7 A-3 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/4*(7*A-3*B)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2
)-1/2*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(1/2)+1/2*(5*A-B)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(
a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3057, 3063, 12, 2861, 211} \begin {gather*} -\frac {(7 A-3 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(5 A-B) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((7*A - 3*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[
2]*a^(3/2)*d) - ((A - B)*Sin[c + d*x])/(2*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + ((5*A - B)*Sin[c
+ d*x])/(2*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 12

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

Rule 211

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

Rule 2861

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 3057

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 3063

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (5 A-B)-a (A-B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int -\frac {a^2 (7 A-3 B)}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{a^3}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-3 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-3 B) \text {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 )}{2 d}\\ &=-\frac {(7 A-3 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 )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac {(5 A-B) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.56, size = 206, normalized size = 1.32 \begin {gather*} \frac {i e^{-\frac {1}{2} i (c+d x)} \left (\frac {2 \left (-1+e^{i (c+d x)}\right ) \left (B \left (1+e^{2 i (c+d x)}\right )-A \left (5+8 e^{i (c+d x)}+5 e^{2 i (c+d x)}\right )\right )}{\left (1+e^{i (c+d x)}\right )^2}-\sqrt {2} (7 A-3 B) \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )}{2 d \sqrt {\cos (c+d x)} (a (1+\cos (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*((2*(-1 + E^(I*(c + d*x)))*(B*(1 + E^((2*I)*(c + d*x))) - A*(5 + 8*E^(I*(c + d*x)) + 5*E^((2*I)*(c + d*
x)))))/(1 + E^(I*(c + d*x)))^2 - Sqrt[2]*(7*A - 3*B)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)
))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])*Cos[(c + d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*Sqrt[Cos[c + d*x]]*(a*
(1 + Cos[c + d*x]))^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(298\) vs. \(2(131)=262\).
time = 0.33, size = 299, normalized size = 1.92

method	result	size
default	\(\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (7 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right )-3 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right )+7 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-3 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-10 A \left (\cos ^{2}\left (d x +c \right )\right )+2 B \left (\cos ^{2}\left (d x +c \right )\right )+2 A \cos \left (d x +c \right )-2 B \cos \left (d x +c \right )+8 A \right )}{4 d \,a^{2} \sin \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/d*(a*(1+cos(d*x+c)))^(1/2)*(7*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c)
)/sin(d*x+c))*cos(d*x+c)-3*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d
*x+c))*cos(d*x+c)+7*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-
3*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-10*A*cos(d*x+c)^2+
2*B*cos(d*x+c)^2+2*A*cos(d*x+c)-2*B*cos(d*x+c)+8*A)/a^2/sin(d*x+c)/(1+cos(d*x+c))/cos(d*x+c)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(3/2)*cos(d*x + c)^(3/2)), x)

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Fricas [A]
time = 0.41, size = 201, normalized size = 1.29 \begin {gather*} -\frac {\sqrt {2} {\left ({\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )\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 - B\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*((7*A - 3*B)*cos(d*x + c)^3 + 2*(7*A - 3*B)*cos(d*x + c)^2 + (7*A - 3*B)*cos(d*x + c))*sqrt(a)*a
rctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d
*x + c))) - 2*((5*A - B)*cos(d*x + c) + 4*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*
cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Integral((A + B*cos(c + d*x))/((a*(cos(c + d*x) + 1))**(3/2)*cos(c + d*x)**(3/2)), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a*cos(c + d*x))^(3/2)),x)

[Out]

int((A + B*cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a*cos(c + d*x))^(3/2)), x)

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