3.134 \(\int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx\)
Optimal. Leaf size=91 \[ -\frac {2^{m+\frac {1}{2}} \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right )}{f \sqrt {\sin (e+f x)+1}} \]
[Out]
-2^(1/2+m)*AppellF1(1/2,-n,1/2-m,3/2,1-sin(f*x+e),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(d*sin(f*x+e))^n/f/(sin(f*x+e
)^n)/(1+sin(f*x+e))^(1/2)
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Rubi [A] time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used =
{2786, 2785, 133} \[ -\frac {2^{m+\frac {1}{2}} \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right )}{f \sqrt {\sin (e+f x)+1}} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^m,x]
[Out]
-((2^(1/2 + m)*AppellF1[1/2, -n, 1/2 - m, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x]*(d*Sin[e +
f*x])^n)/(f*Sin[e + f*x]^n*Sqrt[1 + Sin[e + f*x]]))
Rule 133
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 2785
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Dist[(b*(d
/b)^n*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a - x)^n*(2*a - x)^(m -
1/2))/Sqrt[x], x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]
Rule 2786
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[((d/b
)^IntPart[n]*(d*Sin[e + f*x])^FracPart[n])/(b*Sin[e + f*x])^FracPart[n], Int[(a + b*Sin[e + f*x])^m*(b*Sin[e +
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0] && !Gt
Q[d/b, 0]
Rubi steps
\begin {align*} \int (d \sin (e+f x))^n (1+\sin (e+f x))^m \, dx &=\left (\sin ^{-n}(e+f x) (d \sin (e+f x))^n\right ) \int \sin ^n(e+f x) (1+\sin (e+f x))^m \, dx\\ &=-\frac {\left (\cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n\right ) \operatorname {Subst}\left (\int \frac {(1-x)^n (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1-\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {2^{\frac {1}{2}+m} F_1\left (\frac {1}{2};-n,\frac {1}{2}-m;\frac {3}{2};1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{f \sqrt {1+\sin (e+f x)}}\\ \end {align*}
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Mathematica [B] time = 6.22, size = 2813, normalized size = 30.91 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^m,x]
[Out]
(-3*AppellF1[1/2, -n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f*x]*Si
n[e + f*x]^n*(d*Sin[e + f*x])^n*(1 + Sin[e + f*x])^m)/(f*(Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -n, 1
+ m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*(n*AppellF1[3/2, 1 - n, 1 + m + n,
5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + (1 + m + n)*AppellF1[3/2, -n, 2 + m + n, 5/2,
Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)*((-3*n*AppellF1[1/2, -n
, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f*x]^2*Sin[e + f*x]^(-1 + n
))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e +
Pi/2 - f*x)/2]^2] - 2*(n*AppellF1[3/2, 1 - n, 1 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f
*x)/2]^2] + (1 + m + n)*AppellF1[3/2, -n, 2 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2
]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) + (3*AppellF1[1/2, -n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e
+ Pi/2 - f*x)/2]^2]*Sin[e + f*x]^(1 + n))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -n, 1 + m + n, 3/2
, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*(n*AppellF1[3/2, 1 - n, 1 + m + n, 5/2, Tan[(-e
+ Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + (1 + m + n)*AppellF1[3/2, -n, 2 + m + n, 5/2, Tan[(-e + Pi
/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) - (3*m*AppellF1[1/2, -n, 1 + m + n,
3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f*x]*Sin[e + f*x]^n*Tan[(-e + Pi/2 - f*
x)/2])/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(
-e + Pi/2 - f*x)/2]^2] - 2*(n*AppellF1[3/2, 1 - n, 1 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2
- f*x)/2]^2] + (1 + m + n)*AppellF1[3/2, -n, 2 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*
x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) + (3*Cos[e + f*x]*Sin[e + f*x]^n*(-1/3*(n*AppellF1[3/2, 1 - n, 1 + m +
n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f
*x)/2]) - ((1 + m + n)*AppellF1[3/2, -n, 2 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]
^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/3))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2,
-n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*(n*AppellF1[3/2, 1 - n, 1 + m
+ n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + (1 + m + n)*AppellF1[3/2, -n, 2 + m + n,
5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) - (3*AppellF1[1/2
, -n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f*x]*Sin[e + f*x]^n*(-2
*(n*AppellF1[3/2, 1 - n, 1 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + (1 + m + n
)*AppellF1[3/2, -n, 2 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Sec[(-e + Pi/2 -
f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2] + 3*(-1/3*(n*AppellF1[3/2, 1 - n, 1 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]
^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2]) - ((1 + m + n)*AppellF1[
3/2, -n, 2 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*T
an[(-e + Pi/2 - f*x)/2])/3) - 2*Tan[(-e + Pi/2 - f*x)/2]^2*(n*((-3*(1 + m + n)*AppellF1[5/2, 1 - n, 2 + m + n,
7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x
)/2])/5 + (3*(1 - n)*AppellF1[5/2, 2 - n, 1 + m + n, 7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2
]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/5) + (1 + m + n)*((-3*n*AppellF1[5/2, 1 - n, 2 + m +
n, 7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 -
f*x)/2])/5 - (3*(2 + m + n)*AppellF1[5/2, -n, 3 + m + n, 7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*
x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/5))))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF
1[1/2, -n, 1 + m + n, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*(n*AppellF1[3/2, 1 - n
, 1 + m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + (1 + m + n)*AppellF1[3/2, -n, 2 +
m + n, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)^2)))
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d \sin \left (f x + e\right )\right )^{n} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x, algorithm="fricas")
[Out]
integral((d*sin(f*x + e))^n*(sin(f*x + e) + 1)^m, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x + e\right )\right )^{n} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x, algorithm="giac")
[Out]
integrate((d*sin(f*x + e))^n*(sin(f*x + e) + 1)^m, x)
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maple [F] time = 0.97, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x +e \right )\right )^{n} \left (1+\sin \left (f x +e \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x)
[Out]
int((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x + e\right )\right )^{n} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sin(f*x+e))^n*(1+sin(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate((d*sin(f*x + e))^n*(sin(f*x + e) + 1)^m, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (\sin \left (e+f\,x\right )+1\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sin(e + f*x))^n*(sin(e + f*x) + 1)^m,x)
[Out]
int((d*sin(e + f*x))^n*(sin(e + f*x) + 1)^m, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sin {\left (e + f x \right )}\right )^{n} \left (\sin {\left (e + f x \right )} + 1\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sin(f*x+e))**n*(1+sin(f*x+e))**m,x)
[Out]
Integral((d*sin(e + f*x))**n*(sin(e + f*x) + 1)**m, x)
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