3.7.44 \(\int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{(a+b \cos (e+f x))^{3/2}} \, dx\) [644]

Optimal. Leaf size=213 \[ \frac {2 b (A b-a B) \cos (e+f x) (c \sec (e+f x))^m \sin (e+f x)}{a \left (a^2-b^2\right ) f \sqrt {a+b \cos (e+f x)}}+\frac {2 (c \cos (e+f x))^m (c \sec (e+f x))^m \text {Int}\left (\frac {(c \cos (e+f x))^{-m} \left (\frac {1}{2} c \left (a^2 A+A b^2 (1-2 m)-2 a b B (1-m)\right )-\frac {1}{2} a (A b-a B) c \cos (e+f x)-\frac {1}{2} b (A b-a B) c (3-2 m) \cos ^2(e+f x)\right )}{\sqrt {a+b \cos (e+f x)}},x\right )}{a \left (a^2-b^2\right ) c} \]

[Out]

2*b*(A*b-B*a)*cos(f*x+e)*(c*sec(f*x+e))^m*sin(f*x+e)/a/(a^2-b^2)/f/(a+b*cos(f*x+e))^(1/2)+2*(c*cos(f*x+e))^m*(
c*sec(f*x+e))^m*Unintegrable((1/2*c*(a^2*A+A*b^2*(1-2*m)-2*a*b*B*(1-m))-1/2*a*(A*b-B*a)*c*cos(f*x+e)-1/2*b*(A*
b-B*a)*c*(3-2*m)*cos(f*x+e)^2)/((c*cos(f*x+e))^m)/(a+b*cos(f*x+e))^(1/2),x)/a/(a^2-b^2)/c

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Rubi [A]
time = 0.43, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{(a+b \cos (e+f x))^{3/2}} \, dx \end {gather*}

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*}

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Mathematica [A]
time = 14.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B \cos (e+f x)) (c \sec (e+f x))^m}{(a+b \cos (e+f x))^{3/2}} \, dx \end {gather*}

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 = 0.34, size = 0, normalized size = 0.00 \[\int \frac {\left (A +B \cos \left (f x +e \right )\right ) \left (c \sec \left (f x +e \right )\right )^{m}}{\left (a +b \cos \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(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 [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \sec {\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 {gather*}

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]

Integral((c*sec(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 = 0.00, 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(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)

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {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 )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((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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