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3.5.15 \(\int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx\) [415]

Optimal. Leaf size=74 \[ \frac {\cos (e+f x) \, _2F_1\left (2,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{2 c f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2824, 2746, 70} \begin {gather*} \frac {\cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (2,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{2 c f (2 m+1) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx &=\frac {\cos (e+f x) \int \sec ^3(e+f x) (a+a \sin (e+f x))^{\frac {3}{2}+m} \, dx}{a c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\left (a^2 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+x)^{-\frac {1}{2}+m}}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\cos (e+f x) \, _2F_1\left (2,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{2 c f (1+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 = 15.36, size = 3006, normalized size = 40.62 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-1/8*((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a + a*Sin[e + f*x])^m*(Appel
lF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*
m)*Tan[(-e + Pi/2 - f*x)/4]^2 - (AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/
4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1
 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (2^(1 - 2*m)*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f
*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^4
)^(2*m))/(1 + 2*m)))/(Sqrt[2]*f*(c - c*Sin[e + f*x])^(3/2)*(Cos[Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - P
i/2 + f*x)/2])^3*(-1/8*(m*Cos[(-e + Pi/2 - f*x)/4]*(Cos[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m)*Sin[(-e + Pi/2 - f*
x)/4]*(AppellF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*
x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]^2 - (AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e +
Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^
2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (2^(1 - 2*m)*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-
e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2
 - f*x)/4]^4)^(2*m))/(1 + 2*m)))/Sqrt[2] + ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*((AppellF1[1, -2*m, 2*m, 2, Tan
[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Tan[(-e + Pi/2 -
f*x)/4])/2 + m*AppellF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + P
i/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]^3 + (Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4
]^2*(-1/2*(m*AppellF1[2, 1 - 2*m, 2*m, 3, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + P
i/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) - (m*AppellF1[2, -2*m, 1 + 2*m, 3, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[
(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/2) + (m*AppellF1[1, -2*m, 2*m, 2,
 Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*(Csc[(-e + Pi/2 - f*x)/4]
^2)^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + m*AppellF1[1, -2*m,
 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*(1 - Cot[(-e + Pi
/2 - f*x)/4]^2)^(-1 - 2*m)*(Csc[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (Ap
pellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*(Csc
[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(2*(1 - Cot[(-e + Pi/2 - f*x)/4]^2)
^(2*m)) - (Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*((m*AppellF1[2, 1 - 2*m, 2*m, 3, Cot[
(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*Csc[(-e + Pi/2 - f*x)/4]^2)/2 +
(m*AppellF1[2, -2*m, 1 + 2*m, 3, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x
)/4]*Csc[(-e + Pi/2 - f*x)/4]^2)/2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(
2*m) + (m*AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Csc[(-e + Pi/2 -
f*x)/4]*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sec[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m
))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^
2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + Pi/2
- f*x)/4]^4)^(2*m))/(2^(2*m)*(1 + 2*m)) + (2^(1 - 2*m)*(-1/2*((1 + 2*m)*AppellF1[2 + 2*m, 2*m, 2, 3 + 2*m, (1
- Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f
*x)/4])/(2 + 2*m) - (m*(1 + 2*m)*AppellF1[2 + 2*m, 1 + 2*m, 1, 3 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1
- Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2*(2 + 2*m)))*(-1 + Tan[(-
e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^4)^(2*m))/(1 + 2*m) - (2^(2 - 2*m)*m*AppellF1[1 + 2*m, 2*m
, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*T
an[(-e + Pi/2 - f*x)/4]^3*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^4)^(-1 + 2*m))/(1 +
2*m)))/(8*Sqrt[2])))

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x)

[Out]

int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),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/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(3/2), 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/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2),
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 (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**m/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m/(-c*(sin(e + f*x) - 1))**(3/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(si

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(3/2),x)

[Out]

int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(3/2), x)

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