3.642 \(\int (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^m \, dx\)

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

[Out]

e*AppellF1(1+m,-1/4,-1/4,2+m,(a+b*sin(d*x+c))/(a-b),(a+b*sin(d*x+c))/(a+b))*(a+b*sin(d*x+c))^(1+m)*(e*cos(d*x+
c))^(1/2)/b/d/(1+m)/(1+(-a-b*sin(d*x+c))/(a-b))^(1/4)/(1+(-a-b*sin(d*x+c))/(a+b))^(1/4)

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Rubi [A]  time = 0.10, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2704, 138} \[ \frac {e \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^{m+1} F_1\left (m+1;-\frac {1}{4},-\frac {1}{4};m+2;\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) \sqrt [4]{1-\frac {a+b \sin (c+d x)}{a-b}} \sqrt [4]{1-\frac {a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [F]  time = 5.70, size = 0, normalized size = 0.00 \[ \int (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {e \cos \left (d x + c\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} e \cos \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+b*sin(d*x+c))^m,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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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