3.7.16 \(\int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx\) [616]

Optimal. Leaf size=229 \[ -\frac {b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) (e \cos (c+d x))^{1+p}}{d e (1+p) (2+p) (3+p)}-\frac {a \left (3 b^2+a^2 (2+p)\right ) (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2},\frac {1+p}{2};\frac {3+p}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d e (1+p) (2+p) \sqrt {\sin ^2(c+d x)}}-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)} \]

[Out]

-b*(2*b^2*(2+p)+a^2*(p^2+6*p+11))*(e*cos(d*x+c))^(1+p)/d/e/(3+p)/(p^2+3*p+2)-a*b*(5+p)*(e*cos(d*x+c))^(1+p)*(a
+b*sin(d*x+c))/d/e/(2+p)/(3+p)-b*(e*cos(d*x+c))^(1+p)*(a+b*sin(d*x+c))^2/d/e/(3+p)-a*(3*b^2+a^2*(2+p))*(e*cos(
d*x+c))^(1+p)*hypergeom([1/2, 1/2+1/2*p],[3/2+1/2*p],cos(d*x+c)^2)*sin(d*x+c)/d/e/(1+p)/(2+p)/(sin(d*x+c)^2)^(
1/2)

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Rubi [A]
time = 0.26, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2771, 2941, 2748, 2722} \begin {gather*} -\frac {a \left (a^2 (p+2)+3 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2},\frac {p+1}{2};\frac {p+3}{2};\cos ^2(c+d x)\right )}{d e (p+1) (p+2) \sqrt {\sin ^2(c+d x)}}-\frac {b \left (a^2 \left (p^2+6 p+11\right )+2 b^2 (p+2)\right ) (e \cos (c+d x))^{p+1}}{d e (p+1) (p+2) (p+3)}-\frac {b (a+b \sin (c+d x))^2 (e \cos (c+d x))^{p+1}}{d e (p+3)}-\frac {a b (p+5) (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2) (p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(2*b^2*(2 + p) + a^2*(11 + 6*p + p^2))*(e*Cos[c + d*x])^(1 + p))/(d*e*(1 + p)*(2 + p)*(3 + p))) - (a*(3*b
^2 + a^2*(2 + p))*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[1/2, (1 + p)/2, (3 + p)/2, Cos[c + d*x]^2]*Sin[c
+ d*x])/(d*e*(1 + p)*(2 + p)*Sqrt[Sin[c + d*x]^2]) - (a*b*(5 + p)*(e*Cos[c + d*x])^(1 + p)*(a + b*Sin[c + d*x]
))/(d*e*(2 + p)*(3 + p)) - (b*(e*Cos[c + d*x])^(1 + p)*(a + b*Sin[c + d*x])^2)/(d*e*(3 + p))

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2771

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

Rule 2941

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^p (a+b \sin (c+d x))^3 \, dx &=-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}+\frac {\int (e \cos (c+d x))^p (a+b \sin (c+d x)) \left (2 b^2+a^2 (3+p)+a b (5+p) \sin (c+d x)\right ) \, dx}{3+p}\\ &=-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}+\frac {\int (e \cos (c+d x))^p \left (a (3+p) \left (3 b^2+a^2 (2+p)\right )+b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) \sin (c+d x)\right ) \, dx}{6+5 p+p^2}\\ &=-\frac {b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) (e \cos (c+d x))^{1+p}}{d e (1+p) \left (6+5 p+p^2\right )}-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}+\left (a \left (a^2+\frac {3 b^2}{2+p}\right )\right ) \int (e \cos (c+d x))^p \, dx\\ &=-\frac {b \left (2 b^2 (2+p)+a^2 \left (11+6 p+p^2\right )\right ) (e \cos (c+d x))^{1+p}}{d e (1+p) \left (6+5 p+p^2\right )}-\frac {a \left (a^2+\frac {3 b^2}{2+p}\right ) (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {1}{2},\frac {1+p}{2};\frac {3+p}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d e (1+p) \sqrt {\sin ^2(c+d x)}}-\frac {a b (5+p) (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p) (3+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))^2}{d e (3+p)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 55.71, size = 17703, normalized size = 77.31 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]
time = 0.46, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,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*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^3*(cos(d*x + c)*e)^p, 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*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*sin(d*x + c))*(cos(d*
x + c)*e)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**p*(a+b*sin(d*x+c))**3,x)

[Out]

Integral((e*cos(c + d*x))**p*(a + b*sin(c + d*x))**3, x)

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Giac [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*cos(d*x+c))^p*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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