3.2.43 \(\int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx\) [143]

Optimal. Leaf size=120 \[ -\frac {2 e F_1\left (\frac {5}{2};\frac {1-m}{2},\frac {4-m}{2};\frac {7}{2};\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos ^2(c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{5 a d \sqrt {a+a \sec (c+d x)}} \]

[Out]

-2/5*e*AppellF1(5/2,2-1/2*m,1/2-1/2*m,7/2,-cos(d*x+c),cos(d*x+c))*(1-cos(d*x+c))^(1/2-1/2*m)*cos(d*x+c)^2*(1+c
os(d*x+c))^(1-1/2*m)*(e*sin(d*x+c))^(-1+m)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3961, 2965, 140, 138} \begin {gather*} -\frac {2 e \cos ^2(c+d x) (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{1-\frac {m}{2}} F_1\left (\frac {5}{2};\frac {1-m}{2},\frac {4-m}{2};\frac {7}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{5 a d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*AppellF1[5/2, (1 - m)/2, (4 - m)/2, 7/2, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^((1 - m)/2)*Cos
[c + d*x]^2*(1 + Cos[c + d*x])^(1 - m/2)*(e*Sin[c + d*x])^(-1 + m))/(5*a*d*Sqrt[a + a*Sec[c + d*x]])

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[c^IntPart[n]*((c +
d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 2965

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[g*((g*Cos[e + f*x])^(p - 1)/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e +
f*x])^((p - 1)/2))), Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]],
x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 3961

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Sin[e
+ f*x]^FracPart[m]*((a + b*Csc[e + f*x])^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m]), Int[(g*Cos[e + f*x])^
p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

Rubi steps

\begin {align*} \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx &=\frac {\sqrt {-a-a \cos (c+d x)} \int \frac {(-\cos (c+d x))^{3/2} (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^{3/2}} \, dx}{\sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=-\frac {\left (e (-a-a \cos (c+d x))^{\frac {1}{2}+\frac {1-m}{2}} (-a+a \cos (c+d x))^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int (-x)^{3/2} (-a-a x)^{-\frac {3}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {\left (e (1+\cos (c+d x))^{1-\frac {m}{2}} (-a-a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}+\frac {m}{2}} (-a+a \cos (c+d x))^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int (-x)^{3/2} (1+x)^{-\frac {3}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {\left (e (1-\cos (c+d x))^{\frac {1}{2}-\frac {m}{2}} (1+\cos (c+d x))^{1-\frac {m}{2}} (-a-a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}+\frac {m}{2}} (-a+a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}+\frac {m}{2}} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+m)} (-x)^{3/2} (1+x)^{-\frac {3}{2}+\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=-\frac {2 e F_1\left (\frac {5}{2};\frac {1-m}{2},\frac {4-m}{2};\frac {7}{2};\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos ^2(c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{5 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(484\) vs. \(2(120)=240\).
time = 1.95, size = 484, normalized size = 4.03 \begin {gather*} \frac {4 (3+m) \left (F_1\left (\frac {1+m}{2};-\frac {1}{2},m;\frac {3+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 F_1\left (\frac {1+m}{2};-\frac {1}{2},1+m;\frac {3+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left (-4 (3+m) F_1\left (\frac {1+m}{2};-\frac {1}{2},1+m;\frac {3+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+\left (2 m F_1\left (\frac {3+m}{2};-\frac {1}{2},1+m;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-4 (1+m) F_1\left (\frac {3+m}{2};-\frac {1}{2},2+m;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+F_1\left (\frac {3+m}{2};\frac {1}{2},m;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 F_1\left (\frac {3+m}{2};\frac {1}{2},1+m;\frac {5+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+(3+m) F_1\left (\frac {1+m}{2};-\frac {1}{2},m;\frac {3+m}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right ) (a (1+\sec (c+d x)))^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*(3 + m)*(AppellF1[(1 + m)/2, -1/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*AppellF1[(1 +
m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[(c + d*x)/2]^3*Sin[(c + d*x)/2]*(
e*Sin[c + d*x])^m)/(d*(1 + m)*(-4*(3 + m)*AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan
[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2 + (2*m*AppellF1[(3 + m)/2, -1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Ta
n[(c + d*x)/2]^2] - 4*(1 + m)*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2
]^2] + AppellF1[(3 + m)/2, 1/2, m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*AppellF1[(3 + m)/2,
1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + (3 + m)*AppellF1[(1 +
m)/2, -1/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x]))*(a*(1 + Sec[c + d*x]))^
(3/2))

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sqrt(a*sec(d*x + c) + a)*(e*sin(d*x + c))^m/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))**m/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((e*sin(c + d*x))**m/(a*(sec(c + d*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: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(tan(sageVAR
c/2)^3*t_noste

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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