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

Optimal. Leaf size=154 \[ \frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{11/2}}{11 b^5 d}-\frac{8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}+\frac{2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac{2 (a+b \sin (c+d x))^{15/2}}{15 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{13/2}}{13 b^5 d} \]

[Out]

(2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(7/2))/(7*b^5*d) - (8*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(9/2))/(9*b^5*d
) + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(11/2))/(11*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(13/2))/(13*b^5*d) +
(2*(a + b*Sin[c + d*x])^(15/2))/(15*b^5*d)

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Rubi [A]  time = 0.119945, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2668, 697} \[ \frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{11/2}}{11 b^5 d}-\frac{8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}+\frac{2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac{2 (a+b \sin (c+d x))^{15/2}}{15 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{13/2}}{13 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(7/2))/(7*b^5*d) - (8*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(9/2))/(9*b^5*d
) + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(11/2))/(11*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(13/2))/(13*b^5*d) +
(2*(a + b*Sin[c + d*x])^(15/2))/(15*b^5*d)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^{5/2} \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (a^2-b^2\right )^2 (a+x)^{5/2}-4 \left (a^3-a b^2\right ) (a+x)^{7/2}+2 \left (3 a^2-b^2\right ) (a+x)^{9/2}-4 a (a+x)^{11/2}+(a+x)^{13/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac{8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}+\frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{11/2}}{11 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{13/2}}{13 b^5 d}+\frac{2 (a+b \sin (c+d x))^{15/2}}{15 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.572357, size = 113, normalized size = 0.73 \[ \frac{2 (a+b \sin (c+d x))^{7/2} \left (8190 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^2+6435 \left (a^2-b^2\right )^2+3003 (a+b \sin (c+d x))^4-13860 a (a+b \sin (c+d x))^3-20020 a (a-b) (a+b) (a+b \sin (c+d x))\right )}{45045 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sin[c + d*x])^(7/2)*(6435*(a^2 - b^2)^2 - 20020*a*(a - b)*(a + b)*(a + b*Sin[c + d*x]) + 8190*(3*a^2
- b^2)*(a + b*Sin[c + d*x])^2 - 13860*a*(a + b*Sin[c + d*x])^3 + 3003*(a + b*Sin[c + d*x])^4))/(45045*b^5*d)

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Maple [A]  time = 0.454, size = 126, normalized size = 0.8 \begin{align*}{\frac{6006\,{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+3696\,a{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -2016\,{a}^{2}{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4368\,{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-896\,{a}^{3}b\sin \left ( dx+c \right ) +3584\,a{b}^{3}\sin \left ( dx+c \right ) +256\,{a}^{4}-64\,{a}^{2}{b}^{2}+2496\,{b}^{4}}{45045\,{b}^{5}d} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045/b^5*(a+b*sin(d*x+c))^(7/2)*(3003*b^4*cos(d*x+c)^4+1848*a*b^3*cos(d*x+c)^2*sin(d*x+c)-1008*a^2*b^2*cos(
d*x+c)^2+2184*b^4*cos(d*x+c)^2-448*a^3*b*sin(d*x+c)+1792*a*b^3*sin(d*x+c)+128*a^4-32*a^2*b^2+1248*b^4)/d

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Maxima [A]  time = 0.954272, size = 157, normalized size = 1.02 \begin{align*} \frac{2 \,{\left (3003 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{15}{2}} - 13860 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{13}{2}} a + 8190 \,{\left (3 \, a^{2} - b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}} - 20020 \,{\left (a^{3} - a b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} + 6435 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}\right )}}{45045 \, b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3003*(b*sin(d*x + c) + a)^(15/2) - 13860*(b*sin(d*x + c) + a)^(13/2)*a + 8190*(3*a^2 - b^2)*(b*sin(d*
x + c) + a)^(11/2) - 20020*(a^3 - a*b^2)*(b*sin(d*x + c) + a)^(9/2) + 6435*(a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x
+ c) + a)^(7/2))/(b^5*d)

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Fricas [A]  time = 2.61458, size = 554, normalized size = 3.6 \begin{align*} -\frac{2 \,{\left (7161 \, a b^{6} \cos \left (d x + c\right )^{6} - 128 \, a^{7} + 992 \, a^{5} b^{2} - 6080 \, a^{3} b^{4} - 5536 \, a b^{6} - 7 \,{\left (5 \, a^{3} b^{4} + 79 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \,{\left (3 \, a^{5} b^{2} - 20 \, a^{3} b^{4} - 67 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} +{\left (3003 \, b^{7} \cos \left (d x + c\right )^{6} + 64 \, a^{6} b - 480 \, a^{4} b^{3} - 9088 \, a^{2} b^{5} - 1248 \, b^{7} - 63 \,{\left (71 \, a^{2} b^{5} + 13 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - 8 \,{\left (5 \, a^{4} b^{3} + 718 \, a^{2} b^{5} + 117 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{45045 \, b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45045*(7161*a*b^6*cos(d*x + c)^6 - 128*a^7 + 992*a^5*b^2 - 6080*a^3*b^4 - 5536*a*b^6 - 7*(5*a^3*b^4 + 79*a*
b^6)*cos(d*x + c)^4 + 16*(3*a^5*b^2 - 20*a^3*b^4 - 67*a*b^6)*cos(d*x + c)^2 + (3003*b^7*cos(d*x + c)^6 + 64*a^
6*b - 480*a^4*b^3 - 9088*a^2*b^5 - 1248*b^7 - 63*(71*a^2*b^5 + 13*b^7)*cos(d*x + c)^4 - 8*(5*a^4*b^3 + 718*a^2
*b^5 + 117*b^7)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^5*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)**5*(a+b*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 (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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