3.105 \(\int \frac {(b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=81 \[ \frac {b (2 A+3 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {A b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 81, 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, 3012, 3767, 8} \[ \frac {b (2 A+3 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {A b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

(A*b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(7/2)) + (b*(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 3012

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 3767

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))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {\left (b \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 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (b (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 \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {\left (b (2 A+3 C) \sqrt {b \cos (c+d x)}\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d \sqrt {\cos (c+d x)}}\\ &=\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {b (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 = 0.15, size = 52, normalized size = 0.64 \[ \frac {b \sin (c+d x) \sqrt {b \cos (c+d x)} \left (A \tan ^2(c+d x)+3 (A+C)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

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

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fricas [A]  time = 0.41, size = 50, normalized size = 0.62 \[ \frac {{\left ({\left (2 \, A + 3 \, C\right )} b \cos \left (d x + c\right )^{2} + A b\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)/cos(d*x + c)^(11/2), x)

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maple [A]  time = 0.18, size = 54, normalized size = 0.67 \[ \frac {\left (2 A \left (\cos ^{2}\left (d x +c \right )\right )+3 C \left (\cos ^{2}\left (d x +c \right )\right )+A \right ) \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}} \sin \left (d x +c \right )}{3 d \cos \left (d x +c \right )^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x)

[Out]

1/3/d*(2*A*cos(d*x+c)^2+3*C*cos(d*x+c)^2+A)*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/cos(d*x+c)^(9/2)

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maxima [B]  time = 1.11, size = 355, normalized size = 4.38 \[ \frac {2 \, {\left (\frac {3 \, C b^{\frac {3}{2}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} - \frac {2 \, {\left (3 \, b \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - {\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \sin \left (6 \, d x + 6 \, c\right ) - 3 \, {\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \sin \left (4 \, d x + 4 \, c\right )\right )} A \sqrt {b}}{2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

2/3*(3*C*b^(3/2)*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*
cos(6*d*x + 6*c)*sin(2*d*x + 2*c) + 9*b*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) - (3*b*cos(2*d*x + 2*c) + b)*sin(6*d
*x + 6*c) - 3*(3*b*cos(2*d*x + 2*c) + b)*sin(4*d*x + 4*c))*A*sqrt(b)/(2*(3*cos(4*d*x + 4*c) + 3*cos(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*cos(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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mupad [B]  time = 2.55, size = 218, normalized size = 2.69 \[ \frac {b\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (18\,A\,\sin \left (2\,c+2\,d\,x\right )+12\,A\,\sin \left (4\,c+4\,d\,x\right )+2\,A\,\sin \left (6\,c+6\,d\,x\right )+15\,C\,\sin \left (2\,c+2\,d\,x\right )+12\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )+A\,20{}\mathrm {i}+C\,30{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,30{}\mathrm {i}+A\,\cos \left (4\,c+4\,d\,x\right )\,12{}\mathrm {i}+A\,\cos \left (6\,c+6\,d\,x\right )\,2{}\mathrm {i}+C\,\cos \left (2\,c+2\,d\,x\right )\,45{}\mathrm {i}+C\,\cos \left (4\,c+4\,d\,x\right )\,18{}\mathrm {i}+C\,\cos \left (6\,c+6\,d\,x\right )\,3{}\mathrm {i}\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(3/2))/cos(c + d*x)^(11/2),x)

[Out]

(b*(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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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((b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

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