3.620 \(\int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx\)

Optimal. Leaf size=62 \[ -\frac {4^{-m-1} \cos (e+f x) \, _2F_1\left (\frac {1}{2},m+1;\frac {3}{2};\frac {1-\sin (e+f x)}{4 (\sin (e+f x)+1)}\right )}{f (\sin (e+f x)+1)} \]

[Out]

-4^(-1-m)*cos(f*x+e)*hypergeom([1/2, 1+m],[3/2],1/4*(1-sin(f*x+e))/(1+sin(f*x+e)))/f/(1+sin(f*x+e))

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Rubi [A]  time = 0.11, antiderivative size = 111, normalized size of antiderivative = 1.79, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2788, 132} \[ -\frac {2^{-2 m-1} \cos (e+f x) (\sin (e+f x)+1)^{m-1} \left (\frac {\sin (e+f x)+1}{5 \sin (e+f x)+3}\right )^{\frac {1}{2}-m} (5 \sin (e+f x)+3)^{-m} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};-\frac {1-\sin (e+f x)}{5 \sin (e+f x)+3}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sin[e + f*x])^m*(3 + 5*Sin[e + f*x])^(-1 - m),x]

[Out]

-((2^(-1 - 2*m)*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, -((1 - Sin[e + f*x])/(3 + 5*Sin[e + f*x]))]*
(1 + Sin[e + f*x])^(-1 + m)*((1 + Sin[e + f*x])/(3 + 5*Sin[e + f*x]))^(1/2 - m))/(f*(3 + 5*Sin[e + f*x])^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 2788

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

Rubi steps

\begin {align*} \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx &=\frac {\cos (e+f x) \operatorname {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m} (3+5 x)^{-1-m}}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {2^{-1-2 m} \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};-\frac {1-\sin (e+f x)}{3+5 \sin (e+f x)}\right ) (1+\sin (e+f x))^{-1+m} \left (\frac {1+\sin (e+f x)}{3+5 \sin (e+f x)}\right )^{\frac {1}{2}-m} (3+5 \sin (e+f x))^{-m}}{f}\\ \end {align*}

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Mathematica [C]  time = 1.40, size = 238, normalized size = 3.84 \[ \frac {4^m (\cosh (m \log (4))-\sinh (m \log (4))) (\sin (e+f x)+1)^m (5 \sin (e+f x)+3)^{-m} (\sin (e+f x)+i \cos (e+f x)+1) \left (-\frac {2 \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )+\cos \left (\frac {1}{4} (2 e+2 f x+\pi )\right )}{\sin \left (\frac {1}{4} (2 e+2 f x-\pi )\right )+2 \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}\right )^m \, _2F_1\left (m+1,2 m+1;2 (m+1);\frac {4 \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{2 \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )+\sin \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}\right )}{f (2 m+1) ((2+i) \sin (e+f x)+(-1+2 i) \cos (e+f x)+(2-i))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + Sin[e + f*x])^m*(3 + 5*Sin[e + f*x])^(-1 - m),x]

[Out]

(4^m*Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), (4*Cos[(2*e - Pi + 2*f*x)/4])/(2*Cos[(2*e - Pi + 2*f*x)/4] +
Sin[(2*e - Pi + 2*f*x)/4])]*(1 + Sin[e + f*x])^m*(1 + I*Cos[e + f*x] + Sin[e + f*x])*(-((2*Cos[(2*e - Pi + 2*
f*x)/4] + Cos[(2*e + Pi + 2*f*x)/4])/(2*Cos[(2*e - Pi + 2*f*x)/4] + Sin[(2*e - Pi + 2*f*x)/4])))^m*(Cosh[m*Log
[4]] - Sinh[m*Log[4]]))/(f*(1 + 2*m)*((2 - I) - (1 - 2*I)*Cos[e + f*x] + (2 + I)*Sin[e + f*x])*(3 + 5*Sin[e +
f*x])^m)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (5 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x, algorithm="fricas")

[Out]

integral((5*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (5 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x, algorithm="giac")

[Out]

integrate((5*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, x)

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \left (1+\sin \left (f x +e \right )\right )^{m} \left (3+5 \sin \left (f x +e \right )\right )^{-1-m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x)

[Out]

int((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (5 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x, algorithm="maxima")

[Out]

integrate((5*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (\sin \left (e+f\,x\right )+1\right )}^m}{{\left (5\,\sin \left (e+f\,x\right )+3\right )}^{m+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(e + f*x) + 1)^m/(5*sin(e + f*x) + 3)^(m + 1),x)

[Out]

int((sin(e + f*x) + 1)^m/(5*sin(e + f*x) + 3)^(m + 1), 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((1+sin(f*x+e))**m*(3+5*sin(f*x+e))**(-1-m),x)

[Out]

Timed out

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