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

Optimal. Leaf size=108 \[ \frac {d^2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{12 b \sqrt {d \tan (a+b x)}}+\frac {d \cos (a+b x) \sqrt {d \tan (a+b x)}}{6 b}-\frac {d \cos ^3(a+b x) \sqrt {d \tan (a+b x)}}{3 b} \]

[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

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Rubi [A]
time = 0.10, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2690, 2692, 2694, 2653, 2720} \begin {gather*} \frac {d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \cos ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx &=-\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 F\left (\left .a-\frac {\pi }{4}+b x\right |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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.20, size = 96, normalized size = 0.89 \begin {gather*} -\frac {\cos (a+b x) \left (\sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right )\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)} \end {gather*}

Antiderivative was successfully verified.

[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))

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Maple [A]
time = 0.30, size = 222, normalized size = 2.06

method	result	size
default	\(-\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (\EllipticF \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+2 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}-2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+\cos \left (b x +a \right ) \sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{12 b \sin \left (b x +a \right )^{5}}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/b*(-1+cos(b*x+a))*(EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*sin(b*x+a)*((-1+
cos(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x
+a))^(1/2)+2*cos(b*x+a)^4*2^(1/2)-2*cos(b*x+a)^3*2^(1/2)-cos(b*x+a)^2*2^(1/2)+cos(b*x+a)*2^(1/2))*cos(b*x+a)*(
cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(b*x+a)^5*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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:ext_reduce Error: Bad Argument Typeext_reduce Error: Bad Argument TypeEvaluation time: 10.07Done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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