3.648 \(\int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=311 \[ -\frac {\left (a^2 (m+1)-b^2\right ) (\sin (c+d x)+1)^3 \sec ^4(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^{m+1} \left (\frac {(a+b) (\sin (c+d x)+1)}{(a-b) (\sin (c+d x)-1)}\right )^{\frac {m-2}{2}} \, _2F_1\left (\frac {m}{2},m+1;m+2;-\frac {2 (a+b \sin (c+d x))}{(a-b) (\sin (c+d x)-1)}\right )}{d e^3 m (m+1) (a-b)^3}+\frac {(a (m+2)-2 b) (\sin (c+d x)-1) (\sin (c+d x)+1)^2 \sec ^4(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^{m+1}}{d e^3 m (m+2) (a-b)^2}+\frac {(\sin (c+d x)-1) (\sin (c+d x)+1) \sec ^4(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^{m+1}}{d e^3 (m+2) (a-b)} \]

[Out]

sec(d*x+c)^4*(sin(d*x+c)-1)*(1+sin(d*x+c))*(a+b*sin(d*x+c))^(1+m)/(a-b)/d/e^3/(2+m)/((e*cos(d*x+c))^m)+(-2*b+a
*(2+m))*sec(d*x+c)^4*(sin(d*x+c)-1)*(1+sin(d*x+c))^2*(a+b*sin(d*x+c))^(1+m)/(a-b)^2/d/e^3/m/(2+m)/((e*cos(d*x+
c))^m)-(-b^2+a^2*(1+m))*hypergeom([1+m, 1/2*m],[2+m],-2*(a+b*sin(d*x+c))/(a-b)/(sin(d*x+c)-1))*sec(d*x+c)^4*(1
+sin(d*x+c))^3*((a+b)*(1+sin(d*x+c))/(a-b)/(sin(d*x+c)-1))^(-1+1/2*m)*(a+b*sin(d*x+c))^(1+m)/(a-b)^3/d/e^3/m/(
1+m)/((e*cos(d*x+c))^m)

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Rubi [A]  time = 0.51, antiderivative size = 420, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2700, 2698, 2920, 96, 132} \[ -\frac {b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (m+1,\frac {m+2}{2};m+2;\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) \left (a^2-b^2\right )}+\frac {a (\sin (c+d x)+1) (e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) \left (a^2-b^2\right )}+\frac {a 2^{-m/2} (a m+a+b) (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+2}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (-\frac {m}{2},\frac {m+2}{2};\frac {2-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e m (m+2) (a-b) (a+b)^2}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)} \]

Warning: Unable to verify antiderivative.

[In]

Int[(e*Cos[c + d*x])^(-3 - m)*(a + b*Sin[c + d*x])^m,x]

[Out]

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

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 2698

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(1 - Sin[e + f*x])*(a + b*Sin[e + f*x])^(m + 1)*(-(((a - b)*(1 - Sin[e + f*x]))/((a + b)
*(1 + Sin[e + f*x]))))^(m/2)*Hypergeometric2F1[m + 1, m/2 + 1, m + 2, (2*(a + b*Sin[e + f*x]))/((a + b)*(1 + S
in[e + f*x]))])/(f*(a + b)*(m + 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && EqQ[m + p +
1, 0]

Rule 2700

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[(b*(m + p + 2))/(g^2*(a -
b)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] + Dist[a/(g^2*(a - b)), Int[((g*Cos[
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] && Ne
Q[a^2 - b^2, 0] && ILtQ[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))^{-3-m} (a+b \sin (c+d x))^m \, dx &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}+\frac {a \int \frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2}-\frac {b \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx}{(a-b) e^2 (2+m)}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac {b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac {2+m}{2};2+m;\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac {\left (a (e \cos (c+d x))^{-2-m} (1-\sin (c+d x))^{\frac {2+m}{2}} (1+\sin (c+d x))^{\frac {2+m}{2}}\right ) \operatorname {Subst}\left (\int (1-x)^{-1+\frac {1}{2} (-2-m)} (1+x)^{\frac {1}{2} (-2-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac {b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac {2+m}{2};2+m;\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac {a (e \cos (c+d x))^{-2-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (2+m)}+\frac {\left (a (a+b+a m) (e \cos (c+d x))^{-2-m} (1-\sin (c+d x))^{\frac {2+m}{2}} (1+\sin (c+d x))^{\frac {2+m}{2}}\right ) \operatorname {Subst}\left (\int (1-x)^{\frac {1}{2} (-2-m)} (1+x)^{\frac {1}{2} (-2-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) (a+b) d e (2+m)}\\ &=-\frac {(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac {b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac {2+m}{2};2+m;\frac {2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac {a (e \cos (c+d x))^{-2-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (2+m)}+\frac {2^{-m/2} a (a+b+a m) (e \cos (c+d x))^{-2-m} \, _2F_1\left (-\frac {m}{2},\frac {2+m}{2};\frac {2-m}{2};\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x)) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {2+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e m (2+m)}\\ \end {align*}

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Mathematica [A]  time = 5.05, size = 319, normalized size = 1.03 \[ \frac {\sec ^2(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^m \left (\frac {b (\sin (c+d x)+1) (a+b \sin (c+d x)) \left (\frac {(a+b) (\sin (c+d x)+1)}{(a-b) (\sin (c+d x)-1)}\right )^{m/2} \, _2F_1\left (m+1,\frac {m+2}{2};m+2;-\frac {2 (a+b \sin (c+d x))}{(a-b) (\sin (c+d x)-1)}\right )}{(m+1) (a-b)}+\frac {a (1-\sin (c+d x)) (\sin (c+d x)+1) \left (2^{-m/2} (a m+a+b) \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{m/2} \, _2F_1\left (-\frac {m}{2},\frac {m+2}{2};1-\frac {m}{2};-\frac {(a-b) (\sin (c+d x)-1)}{2 (a+b \sin (c+d x))}\right )-\frac {m (a+b \sin (c+d x))}{\sin (c+d x)-1}\right )}{m (a+b)}-a-b \sin (c+d x)\right )}{d e^3 (m+2) (a-b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-3 - m)*(a + b*Sin[c + d*x])^m,x]

[Out]

(Sec[c + d*x]^2*(a + b*Sin[c + d*x])^m*(-a - b*Sin[c + d*x] + (b*Hypergeometric2F1[1 + m, (2 + m)/2, 2 + m, (-
2*(a + b*Sin[c + d*x]))/((a - b)*(-1 + Sin[c + d*x]))]*(1 + Sin[c + d*x])*(((a + b)*(1 + Sin[c + d*x]))/((a -
b)*(-1 + Sin[c + d*x])))^(m/2)*(a + b*Sin[c + d*x]))/((a - b)*(1 + m)) + (a*(1 - Sin[c + d*x])*(1 + Sin[c + d*
x])*(((a + b + a*m)*Hypergeometric2F1[-1/2*m, (2 + m)/2, 1 - m/2, -1/2*((a - b)*(-1 + Sin[c + d*x]))/(a + b*Si
n[c + d*x])]*(((a + b)*(1 + Sin[c + d*x]))/(a + b*Sin[c + d*x]))^(m/2))/2^(m/2) - (m*(a + b*Sin[c + d*x]))/(-1
+ Sin[c + d*x])))/((a + b)*m)))/((a - b)*d*e^3*(2 + m)*(e*Cos[c + d*x])^m)

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-m)*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(-m - 3)*(b*sin(d*x + c) + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-m)*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m - 3)*(b*sin(d*x + c) + a)^m, x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-3-m} \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-3-m)*(a+b*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-3-m)*(a+b*sin(d*x+c))^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-m)*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m - 3)*(b*sin(d*x + c) + a)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+3}} \,d x \]

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 + 3),x)

[Out]

int((a + b*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 3), 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))**(-3-m)*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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