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

Optimal. Leaf size=212 \[ \frac{2 (c \cos (e+f x))^m (c \sec (e+f x))^m \text{Unintegrable}\left (\frac{(c \cos (e+f x))^{-m} \left (\frac{1}{2} c \left (a^2 A-2 a b B (1-m)+A b^2 (1-2 m)\right )-\frac{1}{2} b c (3-2 m) (A b-a B) \cos ^2(e+f x)-\frac{1}{2} a c (A b-a B) \cos (e+f x)\right )}{\sqrt{a+b \cos (e+f x)}},x\right )}{a c \left (a^2-b^2\right )}+\frac{2 b (A b-a B) \sin (e+f x) \cos (e+f x) (c \sec (e+f x))^m}{a f \left (a^2-b^2\right ) \sqrt{a+b \cos (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.686432, 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 = {} \[ \int \frac{(A+B \cos (e+f x)) (c \sec (e+f x))^m}{(a+b \cos (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \frac{(A+B \cos (e+f x)) (c \sec (e+f x))^m}{(a+b \cos (e+f x))^{3/2}} \, dx &=\left ((c \cos (e+f x))^m (c \sec (e+f x))^m\right ) \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) \cos (e+f x) (c \sec (e+f x))^m \sin (e+f x)}{a \left (a^2-b^2\right ) f \sqrt{a+b \cos (e+f x)}}+\frac{\left (2 (c \cos (e+f x))^m (c \sec (e+f x))^m\right ) \int \frac{(c \cos (e+f x))^{-m} \left (\frac{1}{2} c \left (a^2 A+A b^2 (1-2 m)-2 a b B (1-m)\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 = 10.5023, size = 0, normalized size = 0. \[ \int \frac{(A+B \cos (e+f x)) (c \sec (e+f x))^m}{(a+b \cos (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.716, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+B\cos \left ( fx+e \right ) \right ) \left ( c\sec \left ( fx+e \right ) \right ) ^{m} \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((A+B*cos(f*x+e))*(c*sec(f*x+e))^m/(a+b*cos(f*x+e))^(3/2),x)

[Out]

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

[Out]

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

[Out]

Timed out

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

[Out]

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