3.2.43 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [143]

Optimal. Leaf size=49 \[ \frac {4 (a+a \sin (c+d x))^{11/2}}{11 a^2 d}-\frac {2 (a+a \sin (c+d x))^{13/2}}{13 a^3 d} \]

[Out]

4/11*(a+a*sin(d*x+c))^(11/2)/a^2/d-2/13*(a+a*sin(d*x+c))^(13/2)/a^3/d

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Rubi [A]
time = 0.05, antiderivative size = 49, 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 = {2746, 45} \begin {gather*} \frac {4 (a \sin (c+d x)+a)^{11/2}}{11 a^2 d}-\frac {2 (a \sin (c+d x)+a)^{13/2}}{13 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(4*(a + a*Sin[c + d*x])^(11/2))/(11*a^2*d) - (2*(a + a*Sin[c + d*x])^(13/2))/(13*a^3*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2746

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 44, normalized size = 0.90 \begin {gather*} \frac {2 \left (26 a (a+a \sin (c+d x))^{11/2}-11 (a+a \sin (c+d x))^{13/2}\right )}{143 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(2*(26*a*(a + a*Sin[c + d*x])^(11/2) - 11*(a + a*Sin[c + d*x])^(13/2)))/(143*a^3*d)

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Maple [A]
time = 0.32, size = 31, normalized size = 0.63

method	result	size
default	\(-\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {11}{2}} \left (11 \sin \left (d x +c \right )-15\right )}{143 a^{2} d}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/143/a^2*(a+a*sin(d*x+c))^(11/2)*(11*sin(d*x+c)-15)/d

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Maxima [A]
time = 0.28, size = 38, normalized size = 0.78 \begin {gather*} -\frac {2 \, {\left (11 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 26 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a\right )}}{143 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2/143*(11*(a*sin(d*x + c) + a)^(13/2) - 26*(a*sin(d*x + c) + a)^(11/2)*a)/(a^3*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (41) = 82\).
time = 0.35, size = 102, normalized size = 2.08 \begin {gather*} \frac {2 \, {\left (11 \, a^{3} \cos \left (d x + c\right )^{6} - 68 \, a^{3} \cos \left (d x + c\right )^{4} + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 64 \, a^{3} - 8 \, {\left (5 \, a^{3} \cos \left (d x + c\right )^{4} - 5 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{143 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/143*(11*a^3*cos(d*x + c)^6 - 68*a^3*cos(d*x + c)^4 + 8*a^3*cos(d*x + c)^2 + 64*a^3 - 8*(5*a^3*cos(d*x + c)^4
- 5*a^3*cos(d*x + c)^2 - 8*a^3)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5985 deep

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Giac [A]
time = 7.04, size = 76, normalized size = 1.55 \begin {gather*} -\frac {128 \, \sqrt {2} {\left (11 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 13 \, a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sqrt {a}}{143 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

-128/143*sqrt(2)*(11*a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^13*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 13*a^3*cos(-1
/4*pi + 1/2*d*x + 1/2*c)^11*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + a*sin(c + d*x))^(7/2),x)

[Out]

int(cos(c + d*x)^3*(a + a*sin(c + d*x))^(7/2), x)

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