\(\int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [245]

Optimal result

Integrand size = 21, antiderivative size = 108 \[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{12 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b} \]

[Out]

-1/12*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sec(b*x+a)*sin(2*
b*x+2*a)^(1/2)/b/(d*tan(b*x+a))^(1/2)+1/6*d*cos(b*x+a)*(d*tan(b*x+a))^(1/2)/b-1/3*d*cos(b*x+a)^3*(d*tan(b*x+a)
)^(1/2)/b

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2690, 2692, 2694, 2653, 2720} \[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{12 b \sqrt {d \tan (a+b x)}}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b} \]

[In]

Int[Cos[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]

[Out]

(d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(12*b*Sqrt[d*Tan[a + b*x]]) + (d*Cos[a
+ b*x]*Sqrt[d*Tan[a + b*x]])/(6*b) - (d*Cos[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(3*b)

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2690

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((n - 1)/(a^2*m)), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

Rule 2692

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(a*Sec[e +
f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f*m)), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[2*m, 2*n]

Rule 2694

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b}+\frac {1}{6} d^2 \int \frac {\cos (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx \\ & = \frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b}+\frac {1}{12} d^2 \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx \\ & = \frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b}+\frac {\left (d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{12 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}} \\ & = \frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b}+\frac {\left (d^2 \sec (a+b x) \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{12 \sqrt {d \tan (a+b x)}} \\ & = \frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{12 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.99 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.89 \[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {\cos (a+b x) \left (\sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sqrt {\sec ^2(a+b x)}+\cos (2 (a+b x)) \sqrt {\tan (a+b x)}\right ) (d \tan (a+b x))^{3/2}}{6 b \tan ^{\frac {3}{2}}(a+b x)} \]

[In]

Integrate[Cos[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]

[Out]

-1/6*(Cos[a + b*x]*((-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Sec[a + b*x]^2] +
Cos[2*(a + b*x)]*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(3/2))/(b*Tan[a + b*x]^(3/2))

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.34 (sec) , antiderivative size = 1939, normalized size of antiderivative = 17.95

method	result	size
default	\(\text {Expression too large to display}\)	\(1939\)

[In]

int(cos(b*x+a)^3*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/48/b*sin(b*x+a)*(6*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a)
)^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*
(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-
1/2*I,1/2*2^(1/2))*cos(b*x+a)+6*I*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(1+csc(b*x
+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*cos(b*x+a)-6*I*(1+csc(b*x
+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-
cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))+8*2^(1/2)*cos(b*x+a)^3*sin(b*x+a)-6*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-c
sc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1
/2*I,1/2*2^(1/2))*cos(b*x+a)-6*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-co
t(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(b*x+a)+8*(cot(b*x+a)-csc
(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*
x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)-6*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*
x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*(cot(b*x+a)-csc(b*x
+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a
))^(1/2),1/2-1/2*I,1/2*2^(1/2))+8*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)
-cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))-4*sin(b*x+a)*2^(1/2)*cos(b*x+a)+3*(-
cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-cot(b*x+a
)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-2*cos(b*x+a)
-sin(b*x+a)+csc(b*x+a)+2)/(-1+cos(b*x+a)))*cos(b*x+a)-3*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln((2*
sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+
a))^(1/2)-cot(b*x+a)*cos(b*x+a)+2*cot(b*x+a)+2*cos(b*x+a)+sin(b*x+a)-csc(b*x+a)-2)/(-1+cos(b*x+a)))*cos(b*x+a)
+6*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x
+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos(b*x+a)))*cos(b*x+a)+6*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ar
ctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*
x+a)+3*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-
cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-2*c
os(b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(-1+cos(b*x+a)))-3*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln((2*si
n(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a)
)^(1/2)-cot(b*x+a)*cos(b*x+a)+2*cot(b*x+a)+2*cos(b*x+a)+sin(b*x+a)-csc(b*x+a)-2)/(-1+cos(b*x+a)))+6*(-cos(b*x+
a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/
2)-cos(b*x+a)+1)/(-1+cos(b*x+a)))+6*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)
*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a))))*(d*tan(b*x+a))^(1/2)*d/(-1+co
s(b*x+a))/(cos(b*x+a)+1)*2^(1/2)

Fricas [F]

\[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{3} \,d x } \]

[In]

integrate(cos(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*d*cos(b*x + a)^3*tan(b*x + a), x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)**3*(d*tan(b*x+a))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \cos \left (b x + a\right )^{3} \,d x } \]

[In]

integrate(cos(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*tan(b*x + a))^(3/2)*cos(b*x + a)^3, x)

Giac [F(-2)]

Exception generated. \[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cos(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0]ext_reduce Error: Bad Argument TypeThe choice was done assuming 0=[0,0]
ext_reduce

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\cos \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \]

[In]

int(cos(a + b*x)^3*(d*tan(a + b*x))^(3/2),x)

[Out]

int(cos(a + b*x)^3*(d*tan(a + b*x))^(3/2), x)