\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\) [130]

Optimal result

Integrand size = 35, antiderivative size = 85 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {A \sin (c+d x)}{3 b d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {18, 3091, 3852, 8} \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {(2 A+3 C) \sin (c+d x)}{3 b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 b d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.60 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos (c+d x)} \sin (c+d x) \left (3 (A+C)+A \tan ^2(c+d x)\right )}{3 d (b \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 8.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67

method	result	size
default	\(\frac {\left (2 A \left (\cos ^{2}\left (d x +c \right )\right )+3 C \left (\cos ^{2}\left (d x +c \right )\right )+A \right ) \sin \left (d x +c \right )}{3 b d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(57\)
parts	\(\frac {A \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sin \left (d x +c \right )}{3 d b \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \sin \left (d x +c \right )}{d b \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}}\)	\(79\)
risch	\(\frac {i \left (3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+\left (8 A +9 C \right ) \cos \left (d x +c \right )+i \left (4 A +3 C \right ) \sin \left (d x +c \right )\right )}{3 b \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}\)	\(84\)

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.59 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b^{2} d \cos \left (d x + c\right )^{\frac {7}{2}}} \]

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),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^2*d*cos(d*x + c)^(7/2))

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (73) = 146\).

Time = 0.42 (sec) , antiderivative size = 380, normalized size of antiderivative = 4.47 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, C \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}} + \frac {2 \, {\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} A}{{\left (b \cos \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, b \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, {\left (3 \, b \cos \left (4 \, d x + 4 \, c\right ) + 3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \cos \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (3 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} \cos \left (4 \, d x + 4 \, c\right ) + 6 \, b \cos \left (2 \, d x + 2 \, c\right ) + 6 \, {\left (b \sin \left (4 \, d x + 4 \, c\right ) + b \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + b\right )} \sqrt {b}}\right )}}{3 \, d} \]

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/3*(3*C*sqrt(b)*sin(2*d*x + 2*c)/(b^2*cos(2*d*x + 2*c)^2 + b^2*sin(2*d*x + 2*c)^2 + 2*b^2*cos(2*d*x + 2*c) +
b^2) + 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/((b*cos(6*d*x + 6*c)^2 + 9*b*cos(4*d*x + 4*
c)^2 + 9*b*cos(2*d*x + 2*c)^2 + b*sin(6*d*x + 6*c)^2 + 9*b*sin(4*d*x + 4*c)^2 + 18*b*sin(4*d*x + 4*c)*sin(2*d*
x + 2*c) + 9*b*sin(2*d*x + 2*c)^2 + 2*(3*b*cos(4*d*x + 4*c) + 3*b*cos(2*d*x + 2*c) + b)*cos(6*d*x + 6*c) + 6*(
3*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 6*b*cos(2*d*x + 2*c) + 6*(b*sin(4*d*x + 4*c) + b*sin(2*d*x + 2*c)
)*sin(6*d*x + 6*c) + b)*sqrt(b)))/d

Giac [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.74 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.59 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (18\,A\,\sin \left (2\,c+2\,d\,x\right )+12\,A\,\sin \left (4\,c+4\,d\,x\right )+2\,A\,\sin \left (6\,c+6\,d\,x\right )+15\,C\,\sin \left (2\,c+2\,d\,x\right )+12\,C\,\sin \left (4\,c+4\,d\,x\right )+3\,C\,\sin \left (6\,c+6\,d\,x\right )+A\,20{}\mathrm {i}+C\,30{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,30{}\mathrm {i}+A\,\cos \left (4\,c+4\,d\,x\right )\,12{}\mathrm {i}+A\,\cos \left (6\,c+6\,d\,x\right )\,2{}\mathrm {i}+C\,\cos \left (2\,c+2\,d\,x\right )\,45{}\mathrm {i}+C\,\cos \left (4\,c+4\,d\,x\right )\,18{}\mathrm {i}+C\,\cos \left (6\,c+6\,d\,x\right )\,3{}\mathrm {i}\right )}{3\,b^2\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

[In]

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