3.5.54 \(\int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx\) [454]

Optimal. Leaf size=286 \[ \frac {a (A b-a B) c F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x)}{b \left (a^2-b^2\right ) f}-\frac {(A b-a B) F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^m \cos ^2(e+f x)^{-m/2} \sin (e+f x)}{\left (a^2-b^2\right ) f}-\frac {B (c \cos (e+f x))^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}} \]

[Out]

a*(A*b-B*a)*c*AppellF1(1/2,1/2-1/2*m,1,3/2,sin(f*x+e)^2,-b^2*sin(f*x+e)^2/(a^2-b^2))*(c*cos(f*x+e))^(-1+m)*(co
s(f*x+e)^2)^(1/2-1/2*m)*sin(f*x+e)/b/(a^2-b^2)/f-(A*b-B*a)*AppellF1(1/2,-1/2*m,1,3/2,sin(f*x+e)^2,-b^2*sin(f*x
+e)^2/(a^2-b^2))*(c*cos(f*x+e))^m*sin(f*x+e)/(a^2-b^2)/f/((cos(f*x+e)^2)^(1/2*m))-B*(c*cos(f*x+e))^(1+m)*hyper
geom([1/2, 1/2+1/2*m],[3/2+1/2*m],cos(f*x+e)^2)*sin(f*x+e)/b/c/f/(1+m)/(sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.27, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3081, 2722, 2902, 3268, 440} \begin {gather*} \frac {a c (A b-a B) \sin (e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} (c \cos (e+f x))^{m-1} F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{b f \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (e+f x) \cos ^2(e+f x)^{-m/2} (c \cos (e+f x))^m F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {B \sin (e+f x) (c \cos (e+f x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\cos ^2(e+f x)\right )}{b c f (m+1) \sqrt {\sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*b - a*B)*c*AppellF1[1/2, (1 - m)/2, 1, 3/2, Sin[e + f*x]^2, -((b^2*Sin[e + f*x]^2)/(a^2 - b^2))]*(c*Cos[
e + f*x])^(-1 + m)*(Cos[e + f*x]^2)^((1 - m)/2)*Sin[e + f*x])/(b*(a^2 - b^2)*f) - ((A*b - a*B)*AppellF1[1/2, -
1/2*m, 1, 3/2, Sin[e + f*x]^2, -((b^2*Sin[e + f*x]^2)/(a^2 - b^2))]*(c*Cos[e + f*x])^m*Sin[e + f*x])/((a^2 - b
^2)*f*(Cos[e + f*x]^2)^(m/2)) - (B*(c*Cos[e + f*x])^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[e
+ f*x]^2]*Sin[e + f*x])/(b*c*f*(1 + m)*Sqrt[Sin[e + f*x]^2])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2902

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

Rule 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3268

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, Dist[(-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1
)/2])/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2])), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p,
x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(c \cos (e+f x))^m (A+B \cos (e+f x))}{a+b \cos (e+f x)} \, dx &=\frac {B \int (c \cos (e+f x))^m \, dx}{b}-\frac {(-A b+a B) \int \frac {(c \cos (e+f x))^m}{a+b \cos (e+f x)} \, dx}{b}\\ &=-\frac {B (c \cos (e+f x))^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}}+\frac {(a (A b-a B)) \int \frac {(c \cos (e+f x))^m}{a^2-b^2 \cos ^2(e+f x)} \, dx}{b}-\frac {(A b-a B) \int \frac {(c \cos (e+f x))^{1+m}}{a^2-b^2 \cos ^2(e+f x)} \, dx}{c}\\ &=-\frac {B (c \cos (e+f x))^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}}+\frac {\left (a (A b-a B) c (c \cos (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \cos ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+m)}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (e+f x)\right )}{b f}-\frac {\left ((A b-a B) (c \cos (e+f x))^m \cos ^2(e+f x)^{-m/2}\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{m/2}}{a^2-b^2+b^2 x^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {a (A b-a B) c F_1\left (\frac {1}{2};\frac {1-m}{2},1;\frac {3}{2};\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x)}{b \left (a^2-b^2\right ) f}-\frac {(A b-a B) F_1\left (\frac {1}{2};-\frac {m}{2},1;\frac {3}{2};\sin ^2(e+f x),-\frac {b^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (c \cos (e+f x))^m \cos ^2(e+f x)^{-m/2} \sin (e+f x)}{\left (a^2-b^2\right ) f}-\frac {B (c \cos (e+f x))^{1+m} \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{b c f (1+m) \sqrt {\sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(10482\) vs. \(2(286)=572\).
time = 27.02, size = 10482, normalized size = 36.65 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \frac {\left (c \cos \left (f x +e \right )\right )^{m} \left (A +B \cos \left (f x +e \right )\right )}{a +b \cos \left (f x +e \right )}\, dx\]

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)),x)

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-1,[0,1,0,0]%%%} / %%%{1,[0,0,1,0]%%%}+%%%{-1,[0,0,0,1]%
%%} Error:

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,\cos \left (e+f\,x\right )\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )}{a+b\,\cos \left (e+f\,x\right )} \,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)),x)

[Out]

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

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