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

Optimal. Leaf size=24 \[ \frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \]

[Out]

2/7*(a+b*sin(d*x+c))^(7/2)/b/d

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Rubi [A]  time = 0.04, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2668, 32} \[ \frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sin[c + d*x])^(7/2))/(7*b*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 (c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^{5/2} \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 1.00 \[ \frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sin[c + d*x])^(7/2))/(7*b*d)

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fricas [B]  time = 0.76, size = 77, normalized size = 3.21 \[ -\frac {2 \, {\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{7 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*(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))*sqrt(b*sin(d
*x + c) + a)/(b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88 \[ \frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(a+b*sin(d*x+c))^(7/2)/b/d

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maxima [A]  time = 0.34, size = 20, normalized size = 0.83 \[ \frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}{7 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*sin(d*x + c) + a)^(7/2)/(b*d)

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mupad [B]  time = 5.57, size = 20, normalized size = 0.83 \[ \frac {2\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{7/2}}{7\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a + b*sin(c + d*x))^(7/2))/(7*b*d)

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

[Out]

Timed out

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