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*(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))

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Rubi [A]  time = 0.369708, antiderivative size = 229, normalized size of antiderivative = 1., 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 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])

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 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]

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*}

Mathematica [A]  time = 54.7158, size = 290, normalized size = 1.27 \[ \frac{8 \sec ^2(c+d x)^{p/2} (a+b \sin (c+d x))^3 (e \cos (c+d x))^p \left (\frac{1}{3} a \left (a^2+3 b^2\right ) \tan ^3(c+d x) \, _2F_1\left (\frac{3}{2},\frac{p+4}{2};\frac{5}{2};-\tan ^2(c+d x)\right )-\frac{b \left (3 a^2+b^2\right ) \left ((p+3) \tan ^2(c+d x)+2\right ) \sec ^2(c+d x)^{-\frac{p}{2}-\frac{3}{2}}}{(p+1) (p+3)}-\frac{3 a^2 b \sec ^2(c+d x)^{-\frac{p}{2}-\frac{3}{2}}}{p+3}+a^3 \tan (c+d x) \, _2F_1\left (\frac{1}{2},\frac{p+4}{2};\frac{3}{2};-\tan ^2(c+d x)\right )\right )}{d \left (2 b \left (6 a^2+b^2\right ) \sin (2 (c+d x)) \sqrt{\sec ^2(c+d x)}+8 a^3-12 a b^2 \cos (2 (c+d x))+12 a b^2-b^3 \sin (4 (c+d x)) \sqrt{\sec ^2(c+d x)}\right )} \]

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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Maple [F]  time = 3.557, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{3} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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)