3.142 \(\int \frac{(e \sin (c+d x))^m}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=115 \[ -\frac{2 e \cos (c+d x) (1-\cos (c+d x))^{\frac{1-m}{2}} (\cos (c+d x)+1)^{1-\frac{m}{2}} F_1\left (\frac{3}{2};\frac{1-m}{2},\frac{2-m}{2};\frac{5}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{3 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.32496, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3876, 2886, 135, 133} \[ -\frac{2 e \cos (c+d x) (1-\cos (c+d x))^{\frac{1-m}{2}} (\cos (c+d x)+1)^{1-\frac{m}{2}} F_1\left (\frac{3}{2};\frac{1-m}{2},\frac{2-m}{2};\frac{5}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{3 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3876

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])

Rule 2886

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 135

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 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])

Rubi steps

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

Mathematica [B]  time = 2.0373, size = 277, normalized size = 2.41 \[ \frac{4 (m+3) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{m+1}{2};-\frac{1}{2},m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) (e \sin (c+d x))^m}{d (m+1) \sqrt{a (\sec (c+d x)+1)} \left ((\cos (c+d x)-1) \left (2 (m+1) F_1\left (\frac{m+3}{2};-\frac{1}{2},m+2;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+F_1\left (\frac{m+3}{2};\frac{1}{2},m+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )+(m+3) (\cos (c+d x)+1) F_1\left (\frac{m+1}{2};-\frac{1}{2},m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.208, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\sin \left ( dx+c \right ) \right ) ^{m}{\frac{1}{\sqrt{a+a\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^(1/2),x)

[Out]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt{a \sec \left (d x + c\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))**m/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral((e*sin(c + d*x))**m/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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