3.239 \(\int \cos ^4(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=140 \[ \frac{(3 a-b) (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{8 a^{3/2} f}+\frac{\sin (e+f x) \cos ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 a f}+\frac{(3 a-b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{8 a f} \]

[Out]

((3*a - b)*(a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*a^(3/2)*f) + ((3*a - b)*C
os[e + f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(8*a*f) + (Cos[e + f*x]^3*Sin[e + f*x]*(a + b + b*Tan
[e + f*x]^2)^(3/2))/(4*a*f)

________________________________________________________________________________________

Rubi [A]  time = 0.122573, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4146, 382, 378, 377, 203} \[ \frac{(3 a-b) (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{8 a^{3/2} f}+\frac{\sin (e+f x) \cos ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 a f}+\frac{(3 a-b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{8 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

((3*a - b)*(a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*a^(3/2)*f) + ((3*a - b)*C
os[e + f*x]*Sin[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(8*a*f) + (Cos[e + f*x]^3*Sin[e + f*x]*(a + b + b*Tan
[e + f*x]^2)^(3/2))/(4*a*f)

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^4(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 a f}+\frac{(3 a-b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b+b x^2}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=\frac{(3 a-b) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{8 a f}+\frac{\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 a f}+\frac{((3 a-b) (a+b)) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=\frac{(3 a-b) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{8 a f}+\frac{\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 a f}+\frac{((3 a-b) (a+b)) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{8 a f}\\ &=\frac{(3 a-b) (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{8 a^{3/2} f}+\frac{(3 a-b) \cos (e+f x) \sin (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{8 a f}+\frac{\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 a f}\\ \end{align*}

Mathematica [A]  time = 1.28061, size = 152, normalized size = 1.09 \[ \frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (\sqrt{2} (3 a-b) \sqrt{a+b} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )+\sqrt{a} \sin (e+f x) (a \cos (2 (e+f x))+4 a+b) \sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{a+b}}\right )}{8 a^{3/2} f \sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

(Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*(Sqrt[2]*(3*a - b)*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a +
b]] + Sqrt[a]*(4*a + b + a*Cos[2*(e + f*x)])*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)]*Sin[e + f*x]))/(8*a
^(3/2)*f*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b)])

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Maple [C]  time = 0.39, size = 1713, normalized size = 12.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/8/f/a/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*(-2*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2+
3*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(
a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f
*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b
-b^2)/(a+b)^2)^(1/2))*a^2*sin(f*x+e)+2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(
f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+c
os(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*
b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b*sin(f*x+e)-2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/
2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a
^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b
))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^2*sin(f*x+e)-6
*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a
+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f
*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^
(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*sin(f*x+e)-4*2^(1/2)*(1/(a+b)*(I*cos(f*x+
e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(
1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I
*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b*sin(f*x+e)+2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*
b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(
f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),
-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^
(1/2))*b^2*sin(f*x+e)+2*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2-3*cos(f*x+e)^3*((2*I*a^(1/2)*
b^(1/2)+a-b)/(a+b))^(1/2)*a^2-3*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+3*cos(f*x+e)^2*((2*I*
a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2+3*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-3*cos(f*x+e)*
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2+3*((2*I*a^(
1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2)*cos(f*x+e)*sin(f*x+e)*((b+a*co
s(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*cos(f*x + e)^4, x)

________________________________________________________________________________________

Fricas [A]  time = 1.51429, size = 1355, normalized size = 9.68 \begin{align*} \left [\frac{{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt{-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \,{\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \,{\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \,{\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} - 8 \,{\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \,{\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \,{\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \,{\left (2 \, a^{2} \cos \left (f x + e\right )^{3} +{\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{64 \, a^{2} f}, -\frac{{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \sqrt{a} \arctan \left (\frac{{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \,{\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} +{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \,{\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} -{\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \,{\left (2 \, a^{2} \cos \left (f x + e\right )^{3} +{\left (3 \, a^{2} + a b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, a^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/64*((3*a^2 + 2*a*b - b^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^
4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b +
7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14
*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)
^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(2*a^2*cos(f*x + e)^3 + (3*a^2 + a*b)*cos(f*x + e))*sqrt((a*cos(f*x
+ e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^2*f), -1/32*((3*a^2 + 2*a*b - b^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(
f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) -
4*(2*a^2*cos(f*x + e)^3 + (3*a^2 + a*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)
)/(a^2*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.39471, size = 166, normalized size = 1.19 \begin{align*} -\frac{{\left (\sqrt{-a \sin \left (f x + e\right )^{2} + a + b}{\left (2 \, \sin \left (f x + e\right )^{2} - \frac{5 \, a^{2} + a b}{a^{2}}\right )} \sin \left (f x + e\right ) + \frac{{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \log \left ({\left | -\sqrt{-a} \sin \left (f x + e\right ) + \sqrt{-a \sin \left (f x + e\right )^{2} + a + b} \right |}\right )}{\sqrt{-a} a}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

-1/8*(sqrt(-a*sin(f*x + e)^2 + a + b)*(2*sin(f*x + e)^2 - (5*a^2 + a*b)/a^2)*sin(f*x + e) + (3*a^2 + 2*a*b - b
^2)*log(abs(-sqrt(-a)*sin(f*x + e) + sqrt(-a*sin(f*x + e)^2 + a + b)))/(sqrt(-a)*a))*sgn(cos(f*x + e))/f