3.1.4 \(\int \frac {(a+a \sin (e+f x))^m (A+C \sin ^2(e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [4]

Optimal. Leaf size=123 \[ \frac {(A+C) \cos (e+f x) \, _2F_1\left (1,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3117, 2824, 2746, 70} \begin {gather*} \frac {(A+C) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

((A + C)*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^m)/(f*
(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) - (2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(3 + 2*m)*Sqrt[c -
c*Sin[e + f*x]])

Rule 70

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

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2824

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*
FracPart[m])), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

Rule 3117

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2))/Sqrt[(c_.) + (d_
.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*C*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + 3)*
Sqrt[c + d*Sin[e + f*x]])), x] + Dist[A + C, Int[(a + b*Sin[e + f*x])^m/Sqrt[c + d*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, c, d, e, f, A, C, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 75.01, size = 19244, normalized size = 156.46 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

Result too large to show

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Maple [F]
time = 0.65, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\sqrt {c -c \sin \left (f x +e \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),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((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + A)*(a*sin(f*x + e) + a)^m/sqrt(-c*sin(f*x + e) + c), 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((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(f*x + e)^2 - A - C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**m*(A+C*sin(f*x+e)**2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + C*sin(e + f*x)**2)/sqrt(-c*(sin(e + f*x) - 1)), x)

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Giac [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((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/2),x)

[Out]

int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/2), x)

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