3.122 \(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=79 \[ \frac {(2 A+3 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 79, 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 = {18, 3012, 3767, 8} \[ \frac {(2 A+3 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[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 {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{\sqrt {b \cos (c+d x)}}\\ &=\frac {A \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left ((2 A+3 C) \sqrt {\cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{3 \sqrt {b \cos (c+d x)}}\\ &=\frac {A \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}-\frac {\left ((2 A+3 C) \sqrt {\cos (c+d x)}\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d \sqrt {b \cos (c+d x)}}\\ &=\frac {A \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 51, normalized size = 0.65 \[ \frac {\sin (c+d x) \left (A \tan ^2(c+d x)+3 (A+C)\right )}{3 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 50, normalized size = 0.63 \[ \frac {{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b d \cos \left (d x + c\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/(sqrt(b*cos(d*x + c))*cos(d*x + c)^(7/2)), x)

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maple [A]  time = 0.23, size = 54, normalized size = 0.68 \[ \frac {\sin \left (d x +c \right ) \left (2 A \left (\cos ^{2}\left (d x +c \right )\right )+3 C \left (\cos ^{2}\left (d x +c \right )\right )+A \right )}{3 d \sqrt {b \cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.00, size = 355, normalized size = 4.49 \[ \frac {2 \, {\left (\frac {3 \, C \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, d x + 2 \, c\right ) + b} + \frac {2 \, {\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} A}{{\left (2 \, {\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {b}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.83, size = 220, normalized size = 2.78 \[ \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (18\,A\,\sin \left (2\,c+2\,d\,x\right )+12\,A\,\sin \left (4\,c+4\,d\,x\right )+2\,A\,\sin \left (6\,c+6\,d\,x\right )+15\,C\,\sin \left (2\,c+2\,d\,x\right )+12\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )+A\,20{}\mathrm {i}+C\,30{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,30{}\mathrm {i}+A\,\cos \left (4\,c+4\,d\,x\right )\,12{}\mathrm {i}+A\,\cos \left (6\,c+6\,d\,x\right )\,2{}\mathrm {i}+C\,\cos \left (2\,c+2\,d\,x\right )\,45{}\mathrm {i}+C\,\cos \left (4\,c+4\,d\,x\right )\,18{}\mathrm {i}+C\,\cos \left (6\,c+6\,d\,x\right )\,3{}\mathrm {i}\right )}{3\,b\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((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*si
n(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*b*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((A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2)/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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