3.7.21 \(\int (1+\sin (e+f x))^m (3+4 \sin (e+f x))^{-1-m} \, dx\) [621]

Optimal. Leaf size=64 \[ -\frac {\left (\frac {7}{2}\right )^{-1-m} \cos (e+f x) \, _2F_1\left (\frac {1}{2},1+m;\frac {3}{2};\frac {1-\sin (e+f x)}{7 (1+\sin (e+f x))}\right )}{f (1+\sin (e+f x))} \]

[Out]

-(7/2)^(-1-m)*cos(f*x+e)*hypergeom([1/2, 1+m],[3/2],1/7*(1-sin(f*x+e))/(1+sin(f*x+e)))/f/(1+sin(f*x+e))

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Rubi [A]
time = 0.08, antiderivative size = 122, normalized size of antiderivative = 1.91, 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 = {2867, 134} \begin {gather*} -\frac {2^{m+\frac {1}{2}} 7^{-m-\frac {1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{m-1} \left (\frac {\sin (e+f x)+1}{4 \sin (e+f x)+3}\right )^{\frac {1}{2}-m} (4 \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)}{2 (4 \sin (e+f x)+3)}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Sin[e + f*x])^m*(3 + 4*Sin[e + f*x])^(-1 - m),x]

[Out]

-((2^(1/2 + m)*7^(-1/2 - m)*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, -1/2*(1 - Sin[e + f*x])/(3 + 4*S
in[e + f*x])]*(1 + Sin[e + f*x])^(-1 + m)*((1 + Sin[e + f*x])/(3 + 4*Sin[e + f*x]))^(1/2 - m))/(f*(3 + 4*Sin[e
+ f*x])^m))

Rule 134

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)/((b*e - a*f)*(m + 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)*((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 2867

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+4 \sin (e+f x))^{-1-m} \, dx &=\frac {\cos (e+f x) \text {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m} (3+4 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^{\frac {1}{2}+m} 7^{-\frac {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)}{2 (3+4 \sin (e+f x))}\right ) (1+\sin (e+f x))^{-1+m} \left (\frac {1+\sin (e+f x)}{3+4 \sin (e+f x)}\right )^{\frac {1}{2}-m} (3+4 \sin (e+f x))^{-m}}{f}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 88, normalized size = 1.38 \begin {gather*} -\frac {2\ 7^{-1-m} \cot \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) \, _2F_1\left (\frac {1}{2},1+m;\frac {3}{2};\frac {1}{7} \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right ) (1+\sin (e+f x))^m \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{-m}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sin[e + f*x])^m*(3 + 4*Sin[e + f*x])^(-1 - m),x]

[Out]

(-2*7^(-1 - m)*Cot[(2*e + Pi + 2*f*x)/4]*Hypergeometric2F1[1/2, 1 + m, 3/2, Tan[(2*e - Pi + 2*f*x)/4]^2/7]*(1
+ Sin[e + f*x])^m)/(f*(Sin[(2*e + Pi + 2*f*x)/4]^2)^m)

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \left (1+\sin \left (f x +e \right )\right )^{m} \left (3+4 \sin \left (f x +e \right )\right )^{-1-m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+sin(f*x+e))^m*(3+4*sin(f*x+e))^(-1-m),x)

[Out]

int((1+sin(f*x+e))^m*(3+4*sin(f*x+e))^(-1-m),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((1+sin(f*x+e))^m*(3+4*sin(f*x+e))^(-1-m),x, algorithm="maxima")

[Out]

integrate((4*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, 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((1+sin(f*x+e))^m*(3+4*sin(f*x+e))^(-1-m),x, algorithm="fricas")

[Out]

integral((4*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, 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((1+sin(f*x+e))**m*(3+4*sin(f*x+e))**(-1-m),x)

[Out]

Timed out

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Giac [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((1+sin(f*x+e))^m*(3+4*sin(f*x+e))^(-1-m),x, algorithm="giac")

[Out]

integrate((4*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 \begin {gather*} \int \frac {{\left (\sin \left (e+f\,x\right )+1\right )}^m}{{\left (4\,\sin \left (e+f\,x\right )+3\right )}^{m+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(e + f*x) + 1)^m/(4*sin(e + f*x) + 3)^(m + 1),x)

[Out]

int((sin(e + f*x) + 1)^m/(4*sin(e + f*x) + 3)^(m + 1), x)

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