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]

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

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Rubi [A]  time = 0.12, antiderivative size = 154, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.58, 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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fricas [A]  time = 1.00, size = 224, normalized size = 1.45 \[ -\frac {2 \, {\left (7161 \, a b^{6} \cos \left (d x + c\right )^{6} - 128 \, a^{7} + 992 \, a^{5} b^{2} - 6080 \, a^{3} b^{4} - 5536 \, a b^{6} - 7 \, {\left (5 \, a^{3} b^{4} + 79 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (3 \, a^{5} b^{2} - 20 \, a^{3} b^{4} - 67 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (3003 \, b^{7} \cos \left (d x + c\right )^{6} + 64 \, a^{6} b - 480 \, a^{4} b^{3} - 9088 \, a^{2} b^{5} - 1248 \, b^{7} - 63 \, {\left (71 \, a^{2} b^{5} + 13 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (5 \, a^{4} b^{3} + 718 \, a^{2} b^{5} + 117 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{45045 \, b^{5} d} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{5}\,{d x} \]

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)

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maple [A]  time = 0.64, size = 126, normalized size = 0.82 \[ \frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}} \left (3003 b^{4} \left (\cos ^{4}\left (d x +c \right )\right )+1848 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-1008 a^{2} b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+2184 b^{4} \left (\cos ^{2}\left (d x +c \right )\right )-448 a^{3} b \sin \left (d x +c \right )+1792 a \,b^{3} \sin \left (d x +c \right )+128 a^{4}-32 a^{2} b^{2}+1248 b^{4}\right )}{45045 b^{5} d} \]

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 = 0.33, size = 116, normalized size = 0.75 \[ \frac {2 \, {\left (3003 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 13860 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 8190 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 20020 \, {\left (a^{3} - a b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} + 6435 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}\right )}}{45045 \, b^{5} d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^5\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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