3.2.98 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [198]

Optimal. Leaf size=145 \[ \frac {2 B \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(A-5 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3056, 3061, 2861, 211, 2853, 222} \begin {gather*} \frac {(A-5 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 {2 B \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{3/2} d}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 222

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

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, 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 3056

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 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 2.65, size = 313, normalized size = 2.16 \begin {gather*} \frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\sqrt {2} e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (4 B d x-4 i B \sinh ^{-1}\left (e^{i (c+d x)}\right )-i \sqrt {2} (A-5 B) \log \left (1+e^{i (c+d x)}\right )+4 i B \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )+i \sqrt {2} A \log \left (1-e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )-5 i \sqrt {2} B \log \left (1-e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )}{d \sqrt {1+e^{2 i (c+d x)}}}+\frac {2 (A-B) \sqrt {\cos (c+d x)} \sec \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{d}\right )}{2 (a (1+\cos (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*((Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(4*B*d*x - (
4*I)*B*ArcSinh[E^(I*(c + d*x))] - I*Sqrt[2]*(A - 5*B)*Log[1 + E^(I*(c + d*x))] + (4*I)*B*Log[1 + Sqrt[1 + E^((
2*I)*(c + d*x))]] + I*Sqrt[2]*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - (5*I)*Sqrt[
2]*B*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/(d*Sqrt[1 + E^((2*I)*(c + d*x))]) + (2
*(A - B)*Sqrt[Cos[c + d*x]]*Sec[(c + d*x)/2]*Tan[(c + d*x)/2])/d))/(2*(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. \(297\) vs. \(2(120)=240\).
time = 0.30, size = 298, normalized size = 2.06

method	result	size
default	\(-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (2 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+A \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-5 B \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-2 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}-8 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )-2 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+2 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{5} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} a^{2}}\)	\(298\)

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

[Out]

-1/4/d*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^2*(2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(
d*x+c)^2+A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-5*B*2^(1/2)*sin(d*x+c)*cos(d*x+c)*
arcsin((-1+cos(d*x+c))/sin(d*x+c))-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-8*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)-2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+2*B*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c))/sin(d*x+c)^5/(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)/a^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(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.12, size = 203, normalized size = 1.40 \begin {gather*} -\frac {\sqrt {2} {\left ({\left (A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - 5 \, B\right )} \cos \left (d x + c\right ) + A - 5 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, {\left (B \cos \left (d x + c\right )^{2} + 2 \, B \cos \left (d x + c\right ) + B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{4 \, {\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 {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*((A - 5*B)*cos(d*x + c)^2 + 2*(A - 5*B)*cos(d*x + c) + A - 5*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*co
s(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*sqrt(a*cos(d*x + c) + a)*(A - B)*sqrt(cos(d*x +
c))*sin(d*x + c) + 8*(B*cos(d*x + c)^2 + 2*B*cos(d*x + c) + B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(c
os(d*x + c))/(sqrt(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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))**(3/2),x)

[Out]

Integral((A + B*cos(c + d*x))*sqrt(cos(c + d*x))/(a*(cos(c + d*x) + 1))**(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(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(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 {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\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((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

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

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