3.2.14 \(\int \frac {(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {13}{2}}(c+d x)} \, dx\) [114]

Optimal. Leaf size=85 \[ \frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {b^2 (2 A+3 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/3*A*b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(7/2)+1/3*b^2*(2*A+3*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)
/d/cos(d*x+c)^(3/2)

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Rubi [A]
time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {17, 3091, 3852, 8} \begin {gather*} \frac {b^2 (2 A+3 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {A b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2),x]

[Out]

(A*b^2*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(7/2)) + (b^2*(2*A + 3*C)*Sqrt[b*Cos[c + d*x]]*Sin
[c + d*x])/(3*d*Cos[c + d*x]^(3/2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + 1/2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
 f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx &=\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (b^2 (2 A+3 C) \sqrt {b \cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{3 \sqrt {\cos (c+d x)}}\\ &=\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {\left (b^2 (2 A+3 C) \sqrt {b \cos (c+d x)}\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d \sqrt {\cos (c+d x)}}\\ &=\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {b^2 (2 A+3 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 51, normalized size = 0.60 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} \sin (c+d x) \left (3 (A+C)+A \tan ^2(c+d x)\right )}{3 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*Sin[c + d*x]*(3*(A + C) + A*Tan[c + d*x]^2))/(3*d*Cos[c + d*x]^(7/2))

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Maple [A]
time = 0.25, size = 54, normalized size = 0.64

method result size
default \(\frac {\left (2 A \left (\cos ^{2}\left (d x +c \right )\right )+3 C \left (\cos ^{2}\left (d x +c \right )\right )+A \right ) \sin \left (d x +c \right ) \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{3 d \cos \left (d x +c \right )^{\frac {11}{2}}}\) \(54\)
risch \(\frac {2 i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (3 C \,{\mathrm e}^{4 i \left (d x +c \right )}+6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+6 C \,{\mathrm e}^{2 i \left (d x +c \right )}+2 A +3 C \right )}{3 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\) \(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x,method=_RETURNVERBOSE)

[Out]

1/3/d*(2*A*cos(d*x+c)^2+3*C*cos(d*x+c)^2+A)*sin(d*x+c)*(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(11/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (73) = 146\).
time = 0.64, size = 367, normalized size = 4.32 \begin {gather*} \frac {2 \, {\left (\frac {3 \, C b^{\frac {5}{2}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} - \frac {2 \, {\left (3 \, b^{2} \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - {\left (3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right ) - 3 \, {\left (3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )\right )} A \sqrt {b}}{2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}\right )}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

2/3*(3*C*b^(5/2)*sin(2*d*x + 2*c)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) - 2*(3*b^
2*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) + 9*b^2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) - (3*b^2*cos(2*d*x + 2*c) + b^2)
*sin(6*d*x + 6*c) - 3*(3*b^2*cos(2*d*x + 2*c) + b^2)*sin(4*d*x + 4*c))*A*sqrt(b)/(2*(3*cos(4*d*x + 4*c) + 3*co
s(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*co
s(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x
 + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x +
 2*c) + 1))/d

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Fricas [A]
time = 0.37, size = 54, normalized size = 0.64 \begin {gather*} \frac {{\left ({\left (2 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

1/3*((2*A + 3*C)*b^2*cos(d*x + c)^2 + A*b^2)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^(7/2))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(13/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(13/2), x)

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Mupad [B]
time = 2.36, size = 220, normalized size = 2.59 \begin {gather*} \frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (18\,A\,\sin \left (2\,c+2\,d\,x\right )+12\,A\,\sin \left (4\,c+4\,d\,x\right )+2\,A\,\sin \left (6\,c+6\,d\,x\right )+15\,C\,\sin \left (2\,c+2\,d\,x\right )+12\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )+A\,20{}\mathrm {i}+C\,30{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,30{}\mathrm {i}+A\,\cos \left (4\,c+4\,d\,x\right )\,12{}\mathrm {i}+A\,\cos \left (6\,c+6\,d\,x\right )\,2{}\mathrm {i}+C\,\cos \left (2\,c+2\,d\,x\right )\,45{}\mathrm {i}+C\,\cos \left (4\,c+4\,d\,x\right )\,18{}\mathrm {i}+C\,\cos \left (6\,c+6\,d\,x\right )\,3{}\mathrm {i}\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(13/2),x)

[Out]

(b^2*(b*cos(c + d*x))^(1/2)*(A*20i + C*30i + A*cos(2*c + 2*d*x)*30i + A*cos(4*c + 4*d*x)*12i + A*cos(6*c + 6*d
*x)*2i + C*cos(2*c + 2*d*x)*45i + C*cos(4*c + 4*d*x)*18i + C*cos(6*c + 6*d*x)*3i + 18*A*sin(2*c + 2*d*x) + 12*
A*sin(4*c + 4*d*x) + 2*A*sin(6*c + 6*d*x) + 15*C*sin(2*c + 2*d*x) + 12*C*sin(4*c + 4*d*x) + 3*C*sin(6*c + 6*d*
x)))/(3*d*cos(c + d*x)^(1/2)*(15*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + cos(6*c + 6*d*x) + 10))

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