3.2.55 \(\int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx\) [155]

Optimal. Leaf size=121 \[ \frac {A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(2 A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

A*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+1/3*(2*A+B)*(cos
(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-A*sin(d*x+c)*cos(d*x+c
)^(1/2)/a^2/d/(1+cos(d*x+c))-1/3*(A-B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^2

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Rubi [A]
time = 0.18, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3057, 2827, 2720, 2719} \begin {gather*} \frac {(2 A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {A \sin (c+d x) \sqrt {\cos (c+d x)}}{a^2 d (\cos (c+d x)+1)}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*EllipticE[(c + d*x)/2, 2])/(a^2*d) + ((2*A + B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (A*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) - ((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x
])^2)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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])

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx &=-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (5 A+B)-\frac {1}{2} a (A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a^2 (2 A+B)+\frac {3}{2} a^2 A \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {A \int \sqrt {\cos (c+d x)} \, dx}{2 a^2}+\frac {(2 A+B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac {A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(2 A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.59, size = 815, normalized size = 6.74 \begin {gather*} \frac {i A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{2 (a+a \cos (c+d x))^2}-\frac {4 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (a+a \cos (c+d x))^2 \sqrt {1+\cot ^2(c)}}-\frac {2 B \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (a+a \cos (c+d x))^2 \sqrt {1+\cot ^2(c)}}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {4 A \csc (c)}{d}-\frac {4 A \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {2 (A-B) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(a+a \cos (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*
d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)
]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 +
E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2
*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] +
I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*Cos[c
+ d*x])^2 - (4*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^
2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin
[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2])
- (2*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - Arc
Tan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + (Cos[c/2
+ (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((-4*A*Csc[c])/d - (4*A*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d - (2*Sec[
c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(3*d) - (2*(A - B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])
/(3*d)))/(a + a*Cos[c + d*x])^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(165)=330\).
time = 0.35, size = 350, normalized size = 2.89

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+A -B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2*d*x+1/2*c)^6-4*A*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+6*A*cos
(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),
2^(1/2))-2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))*cos(1/2*d*x+1/2*c)^3-16*A*cos(1/2*d*x+1/2*c)^4-2*B*cos(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d*x+1/2*c)^2+3*B*cos(1
/2*d*x+1/2*c)^2+A-B)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x
+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 318, normalized size = 2.63 \begin {gather*} -\frac {2 \, {\left (3 \, A \cos \left (d x + c\right ) + 4 \, A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (-2 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (2 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-2 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (-i \, \sqrt {2} A \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} A \cos \left (d x + c\right ) - i \, \sqrt {2} A\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} A \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} A \cos \left (d x + c\right ) + i \, \sqrt {2} A\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\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((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(2*(3*A*cos(d*x + c) + 4*A - B)*sqrt(cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(-2*I*A - I*B)*cos(d*x + c)^2
- 2*sqrt(2)*(2*I*A + I*B)*cos(d*x + c) + sqrt(2)*(-2*I*A - I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c)) - (sqrt(2)*(2*I*A + I*B)*cos(d*x + c)^2 - 2*sqrt(2)*(-2*I*A - I*B)*cos(d*x + c) + sqrt(2)*(2*I*A
+ I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(-I*sqrt(2)*A*cos(d*x + c)^2 - 2*I*sqrt(
2)*A*cos(d*x + c) - I*sqrt(2)*A)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c))) + 3*(I*sqrt(2)*A*cos(d*x + c)^2 + 2*I*sqrt(2)*A*cos(d*x + c) + I*sqrt(2)*A)*weierstrassZeta(-4, 0, weiers
trassPInverse(-4, 0, cos(d*x + c) - I*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)] 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))/(a+a*cos(d*x+c))**2/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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