3.128 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=159 \[ -\frac{32 a^2 \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{143 d}-\frac{128 a^3 \cos ^5(c+d x)}{429 d \sqrt{a \sin (c+d x)+a}}-\frac{1024 a^4 \cos ^5(c+d x)}{3003 d (a \sin (c+d x)+a)^{3/2}}-\frac{4096 a^5 \cos ^5(c+d x)}{15015 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d} \]

[Out]

(-4096*a^5*Cos[c + d*x]^5)/(15015*d*(a + a*Sin[c + d*x])^(5/2)) - (1024*a^4*Cos[c + d*x]^5)/(3003*d*(a + a*Sin
[c + d*x])^(3/2)) - (128*a^3*Cos[c + d*x]^5)/(429*d*Sqrt[a + a*Sin[c + d*x]]) - (32*a^2*Cos[c + d*x]^5*Sqrt[a
+ a*Sin[c + d*x]])/(143*d) - (2*a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d)

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Rubi [A]  time = 0.293338, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2674, 2673} \[ -\frac{32 a^2 \cos ^5(c+d x) \sqrt{a \sin (c+d x)+a}}{143 d}-\frac{128 a^3 \cos ^5(c+d x)}{429 d \sqrt{a \sin (c+d x)+a}}-\frac{1024 a^4 \cos ^5(c+d x)}{3003 d (a \sin (c+d x)+a)^{3/2}}-\frac{4096 a^5 \cos ^5(c+d x)}{15015 d (a \sin (c+d x)+a)^{5/2}}-\frac{2 a \cos ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-4096*a^5*Cos[c + d*x]^5)/(15015*d*(a + a*Sin[c + d*x])^(5/2)) - (1024*a^4*Cos[c + d*x]^5)/(3003*d*(a + a*Sin
[c + d*x])^(3/2)) - (128*a^3*Cos[c + d*x]^5)/(429*d*Sqrt[a + a*Sin[c + d*x]]) - (32*a^2*Cos[c + d*x]^5*Sqrt[a
+ a*Sin[c + d*x]])/(143*d) - (2*a*Cos[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(13*d)

Rule 2674

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[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2673

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 - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+a \sin (c+d x))^{5/2} \, dx &=-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{1}{13} (16 a) \int \cos ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac{32 a^2 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{1}{143} \left (192 a^2\right ) \int \cos ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=-\frac{128 a^3 \cos ^5(c+d x)}{429 d \sqrt{a+a \sin (c+d x)}}-\frac{32 a^2 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{1}{429} \left (512 a^3\right ) \int \frac{\cos ^4(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{1024 a^4 \cos ^5(c+d x)}{3003 d (a+a \sin (c+d x))^{3/2}}-\frac{128 a^3 \cos ^5(c+d x)}{429 d \sqrt{a+a \sin (c+d x)}}-\frac{32 a^2 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}+\frac{\left (2048 a^4\right ) \int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{3003}\\ &=-\frac{4096 a^5 \cos ^5(c+d x)}{15015 d (a+a \sin (c+d x))^{5/2}}-\frac{1024 a^4 \cos ^5(c+d x)}{3003 d (a+a \sin (c+d x))^{3/2}}-\frac{128 a^3 \cos ^5(c+d x)}{429 d \sqrt{a+a \sin (c+d x)}}-\frac{32 a^2 \cos ^5(c+d x) \sqrt{a+a \sin (c+d x)}}{143 d}-\frac{2 a \cos ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{13 d}\\ \end{align*}

Mathematica [A]  time = 0.309193, size = 79, normalized size = 0.5 \[ -\frac{2 \left (1155 \sin ^4(c+d x)+6300 \sin ^3(c+d x)+14210 \sin ^2(c+d x)+16700 \sin (c+d x)+9683\right ) \cos ^5(c+d x) (a (\sin (c+d x)+1))^{5/2}}{15015 d (\sin (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-2*Cos[c + d*x]^5*(a*(1 + Sin[c + d*x]))^(5/2)*(9683 + 16700*Sin[c + d*x] + 14210*Sin[c + d*x]^2 + 6300*Sin[c
+ d*x]^3 + 1155*Sin[c + d*x]^4))/(15015*d*(1 + Sin[c + d*x])^5)

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Maple [A]  time = 0.112, size = 87, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ){a}^{3} \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3} \left ( 1155\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+6300\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+14210\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}+16700\,\sin \left ( dx+c \right ) +9683 \right ) }{15015\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/15015*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)^3*(1155*sin(d*x+c)^4+6300*sin(d*x+c)^3+14210*sin(d*x+c)^2+16700*sin(
d*x+c)+9683)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*cos(d*x + c)^4, x)

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Fricas [A]  time = 1.68916, size = 595, normalized size = 3.74 \begin{align*} \frac{2 \,{\left (1155 \, a^{2} \cos \left (d x + c\right )^{7} - 2835 \, a^{2} \cos \left (d x + c\right )^{6} - 6230 \, a^{2} \cos \left (d x + c\right )^{5} + 320 \, a^{2} \cos \left (d x + c\right )^{4} - 512 \, a^{2} \cos \left (d x + c\right )^{3} + 1024 \, a^{2} \cos \left (d x + c\right )^{2} - 4096 \, a^{2} \cos \left (d x + c\right ) - 8192 \, a^{2} -{\left (1155 \, a^{2} \cos \left (d x + c\right )^{6} + 3990 \, a^{2} \cos \left (d x + c\right )^{5} - 2240 \, a^{2} \cos \left (d x + c\right )^{4} - 2560 \, a^{2} \cos \left (d x + c\right )^{3} - 3072 \, a^{2} \cos \left (d x + c\right )^{2} - 4096 \, a^{2} \cos \left (d x + c\right ) - 8192 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{15015 \,{\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/15015*(1155*a^2*cos(d*x + c)^7 - 2835*a^2*cos(d*x + c)^6 - 6230*a^2*cos(d*x + c)^5 + 320*a^2*cos(d*x + c)^4
- 512*a^2*cos(d*x + c)^3 + 1024*a^2*cos(d*x + c)^2 - 4096*a^2*cos(d*x + c) - 8192*a^2 - (1155*a^2*cos(d*x + c)
^6 + 3990*a^2*cos(d*x + c)^5 - 2240*a^2*cos(d*x + c)^4 - 2560*a^2*cos(d*x + c)^3 - 3072*a^2*cos(d*x + c)^2 - 4
096*a^2*cos(d*x + c) - 8192*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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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(cos(d*x+c)**4*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)*cos(d*x + c)^4, x)