3.649 \(\int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=201 \[ \frac {a 2^{\frac {1}{2}-\frac {m}{2}} (e \cos (c+d x))^{-m-1} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+1}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2} (-m-1),\frac {m+1}{2};\frac {1-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e (m+1) \left (a^2-b^2\right )}-\frac {(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1}}{d e (m+1) (a-b)} \]

[Out]

-(e*cos(d*x+c))^(-1-m)*(a+b*sin(d*x+c))^(1+m)/(a-b)/d/e/(1+m)+2^(1/2-1/2*m)*a*(e*cos(d*x+c))^(-1-m)*hypergeom(
[-1/2-1/2*m, 1/2+1/2*m],[1/2-1/2*m],1/2*(a-b)*(1-sin(d*x+c))/(a+b*sin(d*x+c)))*((a+b)*(1+sin(d*x+c))/(a+b*sin(
d*x+c)))^(1/2+1/2*m)*(a+b*sin(d*x+c))^(1+m)/(a^2-b^2)/d/e/(1+m)

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Rubi [A]  time = 0.29, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2699, 2920, 132} \[ \frac {a 2^{\frac {1}{2}-\frac {m}{2}} (e \cos (c+d x))^{-m-1} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+1}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2} (-m-1),\frac {m+1}{2};\frac {1-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e (m+1) \left (a^2-b^2\right )}-\frac {(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1}}{d e (m+1) (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(-2 - m)*(a + b*Sin[c + d*x])^m,x]

[Out]

-(((e*Cos[c + d*x])^(-1 - m)*(a + b*Sin[c + d*x])^(1 + m))/((a - b)*d*e*(1 + m))) + (2^(1/2 - m/2)*a*(e*Cos[c
+ d*x])^(-1 - m)*Hypergeometric2F1[(-1 - m)/2, (1 + m)/2, (1 - m)/2, ((a - b)*(1 - Sin[c + d*x]))/(2*(a + b*Si
n[c + d*x]))]*(((a + b)*(1 + Sin[c + d*x]))/(a + b*Sin[c + d*x]))^((1 + m)/2)*(a + b*Sin[c + d*x])^(1 + m))/((
a^2 - b^2)*d*e*(1 + m))

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rule 2699

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a - b)*(p + 1)), x] + Dist[a/(g^2*(a - b)), Int[((g*Co
s[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m)/(1 - Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && N
eQ[a^2 - b^2, 0] && EqQ[m + p + 2, 0]

Rule 2920

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^m*g*(g*Cos[e + f*x])^(p - 1))/(f*(1 + Sin[e + f*x])^((p - 1)/2)*(1 -
Sin[e + f*x])^((p - 1)/2)), Subst[Int[(1 + (b*x)/a)^(m + (p - 1)/2)*(1 - (b*x)/a)^((p - 1)/2)*(c + d*x)^n, x],
x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx &=-\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {a \int \frac {(e \cos (c+d x))^{-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2}\\ &=-\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {\left (a (e \cos (c+d x))^{-1-m} (1-\sin (c+d x))^{\frac {1+m}{2}} (1+\sin (c+d x))^{\frac {1+m}{2}}\right ) \operatorname {Subst}\left (\int (1-x)^{-1+\frac {1}{2} (-1-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e}\\ &=-\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^{-1-m} \, _2F_1\left (\frac {1}{2} (-1-m),\frac {1+m}{2};\frac {1-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.96, size = 168, normalized size = 0.84 \[ -\frac {2^{\frac {1}{2} (-m-1)} (e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^{m+1} \left (2^{\frac {m+1}{2}} (a+b)-2 a \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+1}{2}} \, _2F_1\left (\frac {1}{2} (-m-1),\frac {m+1}{2};\frac {1-m}{2};-\frac {(a-b) (\sin (c+d x)-1)}{2 (a+b \sin (c+d x))}\right )\right )}{d e (m+1) (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-2 - m)*(a + b*Sin[c + d*x])^m,x]

[Out]

-((2^((-1 - m)/2)*(e*Cos[c + d*x])^(-1 - m)*(a + b*Sin[c + d*x])^(1 + m)*(2^((1 + m)/2)*(a + b) - 2*a*Hypergeo
metric2F1[(-1 - m)/2, (1 + m)/2, (1 - m)/2, -1/2*((a - b)*(-1 + Sin[c + d*x]))/(a + b*Sin[c + d*x])]*(((a + b)
*(1 + Sin[c + d*x]))/(a + b*Sin[c + d*x]))^((1 + m)/2)))/((a - b)*(a + b)*d*e*(1 + m)))

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-2-m)*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(-m - 2)*(b*sin(d*x + c) + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-2-m)*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m - 2)*(b*sin(d*x + c) + a)^m, x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-2-m} \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-2-m)*(a+b*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-2-m)*(a+b*sin(d*x+c))^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-2-m)*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m - 2)*(b*sin(d*x + c) + a)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 2),x)

[Out]

int((a + b*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(-2-m)*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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