\(\int \frac {(b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [104]

Optimal result

Integrand size = 35, antiderivative size = 80 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {b (A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 3091, 3855} \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {b (A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b \sin (c+d x) \sqrt {b \cos (c+d x)}}{2 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[((b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

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

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{\sqrt {\cos (c+d x)}} \\ & = \frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (b (A+2 C) \sqrt {b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{2 \sqrt {\cos (c+d x)}} \\ & = \frac {b (A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.74 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{3/2} \left ((A+2 C) \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+A \sin (c+d x)\right )}{2 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

Integrate[((b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

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

Maple [A] (verified)

Time = 9.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45

method	result	size
default	\(\frac {b \left (-A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-4 C \left (\cos ^{2}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+A \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(116\)
parts	\(\frac {A b \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 C \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) b \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(128\)
risch	\(-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, A \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(146\)

[In]

int((cos(d*x+c)*b)^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/2*b/d*(-A*cos(d*x+c)^2*ln(-cot(d*x+c)+csc(d*x+c)-1)+A*cos(d*x+c)^2*ln(-cot(d*x+c)+csc(d*x+c)+1)-4*C*cos(d*x+
c)^2*arctanh(cot(d*x+c)-csc(d*x+c))+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 = 0.29 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.70 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\left [\frac {{\left (A + 2 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{3}}, -\frac {{\left (A + 2 \, C\right )} \sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} A b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{3}}\right ] \]

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

[1/4*((A + 2*C)*b^(3/2)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x +
c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*A*b*sqrt(cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^3), -1/2*((A + 2*C)*sqrt(-b)*b*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(
cos(d*x + c))))*cos(d*x + c)^3 - sqrt(b*cos(d*x + c))*A*b*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 761 vs. \(2 (68) = 136\).

Time = 0.48 (sec) , antiderivative size = 761, normalized size of antiderivative = 9.51 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

1/4*(2*(b*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b*log(cos(d*x + c)^2 + sin(d*x + c)^2 -
2*sin(d*x + c) + 1))*C*sqrt(b) - (4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c))) - 4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c))) - (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4*d*x + 4*c)^2 + 4*b*sin(4*d*x + 4*c)*s
in(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c
) + b)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2
*d*x + 2*c)^2 + b*sin(4*d*x + 4*c)^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b
*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c))) + 1) - 4*(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c) + b)*sin(3/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c))) + 4*(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c) + b)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c))))*A*sqrt(b)/(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2
*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) +
1))/d

Giac [F]

\[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \]

[In]

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

[Out]

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