3.644 \(\int \frac{(a+b \sin (c+d x))^m}{\sqrt{e \cos (c+d x)}} \, dx\)

Optimal. Leaf size=134 \[ \frac{e \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{3/4} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{3/4} (a+b \sin (c+d x))^{m+1} F_1\left (m+1;\frac{3}{4},\frac{3}{4};m+2;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) (e \cos (c+d x))^{3/2}} \]

[Out]

(e*AppellF1[1 + m, 3/4, 3/4, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*(a + b*Sin[c +
d*x])^(1 + m)*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/4)*(1 - (a + b*Sin[c + d*x])/(a + b))^(3/4))/(b*d*(1 + m)
*(e*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.0943236, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2704, 138} \[ \frac{e \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{3/4} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{3/4} (a+b \sin (c+d x))^{m+1} F_1\left (m+1;\frac{3}{4},\frac{3}{4};m+2;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*AppellF1[1 + m, 3/4, 3/4, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*(a + b*Sin[c +
d*x])^(1 + m)*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/4)*(1 - (a + b*Sin[c + d*x])/(a + b))^(3/4))/(b*d*(1 + m)
*(e*Cos[c + d*x])^(3/2))

Rule 2704

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

Rule 138

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

Rubi steps

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

Mathematica [F]  time = 1.75679, size = 0, normalized size = 0. \[ \int \frac{(a+b \sin (c+d x))^m}{\sqrt{e \cos (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\sin \left ( dx+c \right ) \right ) ^{m}{\frac{1}{\sqrt{e\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sin(c + d*x))**m/sqrt(e*cos(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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