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

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

[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 = 0.37, 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 = {2692, 2862, 2669, 2643} \[ -\frac {a \left (a^2 (p+2)+3 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {1}{2},\frac {p+1}{2};\frac {p+3}{2};\cos ^2(c+d x)\right )}{d e (p+1) (p+2) \sqrt {\sin ^2(c+d x)}}-\frac {b \left (a^2 \left (p^2+6 p+11\right )+2 b^2 (p+2)\right ) (e \cos (c+d x))^{p+1}}{d e (p+1) (p+2) (p+3)}-\frac {b (a+b \sin (c+d x))^2 (e \cos (c+d x))^{p+1}}{d e (p+3)}-\frac {a b (p+5) (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2) (p+3)} \]

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 2692

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 2862

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] && Gt
Q[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 [A]  time = 54.88, size = 290, normalized size = 1.27 \[ \frac {8 \sec ^2(c+d x)^{p/2} (a+b \sin (c+d x))^3 (e \cos (c+d x))^p \left (a^3 \tan (c+d x) \, _2F_1\left (\frac {1}{2},\frac {p+4}{2};\frac {3}{2};-\tan ^2(c+d x)\right )+\frac {1}{3} a \left (a^2+3 b^2\right ) \tan ^3(c+d x) \, _2F_1\left (\frac {3}{2},\frac {p+4}{2};\frac {5}{2};-\tan ^2(c+d x)\right )-\frac {b \left (3 a^2+b^2\right ) \left ((p+3) \tan ^2(c+d x)+2\right ) \sec ^2(c+d x)^{-\frac {p}{2}-\frac {3}{2}}}{(p+1) (p+3)}-\frac {3 a^2 b \sec ^2(c+d x)^{-\frac {p}{2}-\frac {3}{2}}}{p+3}\right )}{d \left (8 a^3+2 b \left (6 a^2+b^2\right ) \sin (2 (c+d x)) \sqrt {\sec ^2(c+d x)}-12 a b^2 \cos (2 (c+d x))+12 a b^2-b^3 \sin (4 (c+d x)) \sqrt {\sec ^2(c+d x)}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*(e*Cos[c + d*x])^p*(Sec[c + d*x]^2)^(p/2)*(a + b*Sin[c + d*x])^3*((-3*a^2*b*(Sec[c + d*x]^2)^(-3/2 - p/2))/
(3 + p) + a^3*Hypergeometric2F1[1/2, (4 + p)/2, 3/2, -Tan[c + d*x]^2]*Tan[c + d*x] + (a*(a^2 + 3*b^2)*Hypergeo
metric2F1[3/2, (4 + p)/2, 5/2, -Tan[c + d*x]^2]*Tan[c + d*x]^3)/3 - (b*(3*a^2 + b^2)*(Sec[c + d*x]^2)^(-3/2 -
p/2)*(2 + (3 + p)*Tan[c + d*x]^2))/((1 + p)*(3 + p))))/(d*(8*a^3 + 12*a*b^2 - 12*a*b^2*Cos[2*(c + d*x)] + 2*b*
(6*a^2 + b^2)*Sqrt[Sec[c + d*x]^2]*Sin[2*(c + d*x)] - b^3*Sqrt[Sec[c + d*x]^2]*Sin[4*(c + d*x)]))

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{p}, x\right ) \]

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))*(e*cos(
d*x + c))^p, x)

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

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*(e*cos(d*x + c))^p, x)

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maple [F]  time = 6.00, 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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \]

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*(e*cos(d*x + c))^p, x)

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

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

[Out]

Timed out

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