3.458 \(\int \frac{(c \cos (e+f x))^m (A+B \cos (e+f x))}{(a+b \cos (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=190 \[ \frac{2 \text{Unintegrable}\left (\frac{(c \cos (e+f x))^m \left (-\frac{1}{2} b c (2 m+3) (A b-a B) \cos ^2(e+f x)-\frac{1}{2} a c (A b-a B) \cos (e+f x)+\frac{1}{2} c \left (2 b \left (m+\frac{1}{2}\right ) (A b-a B)+a (a A-b B)\right )\right )}{\sqrt{a+b \cos (e+f x)}},x\right )}{a c \left (a^2-b^2\right )}+\frac{2 b (A b-a B) \sin (e+f x) (c \cos (e+f x))^{m+1}}{a c f \left (a^2-b^2\right ) \sqrt{a+b \cos (e+f x)}} \]

[Out]

(2*b*(A*b - a*B)*(c*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(a*(a^2 - b^2)*c*f*Sqrt[a + b*Cos[e + f*x]]) + (2*Unin
tegrable[((c*Cos[e + f*x])^m*((c*(a*(a*A - b*B) + 2*b*(A*b - a*B)*(1/2 + m)))/2 - (a*(A*b - a*B)*c*Cos[e + f*x
])/2 - (b*(A*b - a*B)*c*(3 + 2*m)*Cos[e + f*x]^2)/2))/Sqrt[a + b*Cos[e + f*x]], x])/(a*(a^2 - b^2)*c)

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Rubi [A]  time = 0.499744, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{(c \cos (e+f x))^m (A+B \cos (e+f x))}{(a+b \cos (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Int[((c*Cos[e + f*x])^m*(A + B*Cos[e + f*x]))/(a + b*Cos[e + f*x])^(3/2),x]

[Out]

(2*b*(A*b - a*B)*(c*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(a*(a^2 - b^2)*c*f*Sqrt[a + b*Cos[e + f*x]]) + (2*Defe
r[Int][((c*Cos[e + f*x])^m*((c*(a*(a*A - b*B) + 2*b*(A*b - a*B)*(1/2 + m)))/2 - (a*(A*b - a*B)*c*Cos[e + f*x])
/2 - (b*(A*b - a*B)*c*(3 + 2*m)*Cos[e + f*x]^2)/2))/Sqrt[a + b*Cos[e + f*x]], x])/(a*(a^2 - b^2)*c)

Rubi steps

\begin{align*} \int \frac{(c \cos (e+f x))^m (A+B \cos (e+f x))}{(a+b \cos (e+f x))^{3/2}} \, dx &=\frac{2 b (A b-a B) (c \cos (e+f x))^{1+m} \sin (e+f x)}{a \left (a^2-b^2\right ) c f \sqrt{a+b \cos (e+f x)}}+\frac{2 \int \frac{(c \cos (e+f x))^m \left (\frac{1}{2} c \left (a (a A-b B)+2 b (A b-a B) \left (\frac{1}{2}+m\right )\right )-\frac{1}{2} a (A b-a B) c \cos (e+f x)-\frac{1}{2} b (A b-a B) c (3+2 m) \cos ^2(e+f x)\right )}{\sqrt{a+b \cos (e+f x)}} \, dx}{a \left (a^2-b^2\right ) c}\\ \end{align*}

Mathematica [A]  time = 10.5573, size = 0, normalized size = 0. \[ \int \frac{(c \cos (e+f x))^m (A+B \cos (e+f x))}{(a+b \cos (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((c*Cos[e + f*x])^m*(A + B*Cos[e + f*x]))/(a + b*Cos[e + f*x])^(3/2),x]

[Out]

Integrate[((c*Cos[e + f*x])^m*(A + B*Cos[e + f*x]))/(a + b*Cos[e + f*x])^(3/2), x]

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Maple [A]  time = 0.423, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c\cos \left ( fx+e \right ) \right ) ^{m} \left ( A+B\cos \left ( fx+e \right ) \right ) \left ( a+b\cos \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*cos(f*x+e))^m*(A+B*cos(f*x+e))/(a+b*cos(f*x+e))^(3/2),x)

[Out]

int((c*cos(f*x+e))^m*(A+B*cos(f*x+e))/(a+b*cos(f*x+e))^(3/2),x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \cos \left (f x + e\right )\right )^{m}}{{\left (b \cos \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(f*x + e) + A)*(c*cos(f*x + e))^m/(b*cos(f*x + e) + a)^(3/2), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (f x + e\right ) + A\right )} \sqrt{b \cos \left (f x + e\right ) + a} \left (c \cos \left (f x + e\right )\right )^{m}}{b^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(f*x + e) + A)*sqrt(b*cos(f*x + e) + a)*(c*cos(f*x + e))^m/(b^2*cos(f*x + e)^2 + 2*a*b*cos(f*x
+ e) + a^2), x)

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \cos{\left (e + f x \right )}\right )^{m} \left (A + B \cos{\left (e + f x \right )}\right )}{\left (a + b \cos{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*cos(f*x+e))**m*(A+B*cos(f*x+e))/(a+b*cos(f*x+e))**(3/2),x)

[Out]

Integral((c*cos(e + f*x))**m*(A + B*cos(e + f*x))/(a + b*cos(e + f*x))**(3/2), x)

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (f x + e\right ) + A\right )} \left (c \cos \left (f x + e\right )\right )^{m}}{{\left (b \cos \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(f*x + e) + A)*(c*cos(f*x + e))^m/(b*cos(f*x + e) + a)^(3/2), x)