3.453 \(\int \cos ^{\frac {3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \, dx\)

Optimal. Leaf size=371 \[ \frac {2 \left (a^2-b^2+c^2\right ) \cos ^{\frac {3}{2}}(d+e x) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2} F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e (a \cos (d+e x)+b+c \sin (d+e x))^2}+\frac {8 b \cos ^{\frac {3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e (a \cos (d+e x)+b+c \sin (d+e x)) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}-\frac {2 \cos ^{\frac {3}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{3 e (a \cos (d+e x)+b+c \sin (d+e x))} \]

[Out]

-2/3*cos(e*x+d)^(3/2)*(c*cos(e*x+d)-a*sin(e*x+d))*(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/e/(b+a*cos(e*x+d)+c*sin(
e*x+d))+8/3*b*cos(e*x+d)^(3/2)*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(a,c))
*EllipticE(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^(1/2))*(a+b*sec(e*
x+d)+c*tan(e*x+d))^(3/2)/e/(b+a*cos(e*x+d)+c*sin(e*x+d))/((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(
1/2)+2/3*(a^2-b^2+c^2)*cos(e*x+d)^(3/2)*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arc
tan(a,c))*EllipticF(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^(1/2))*((
b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/e/(b+a*cos(e*x+d)+
c*sin(e*x+d))^2

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Rubi [A]  time = 0.39, antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3163, 3120, 3149, 3119, 2653, 3127, 2661} \[ \frac {2 \left (a^2-b^2+c^2\right ) \cos ^{\frac {3}{2}}(d+e x) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2} F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e (a \cos (d+e x)+b+c \sin (d+e x))^2}+\frac {8 b \cos ^{\frac {3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e (a \cos (d+e x)+b+c \sin (d+e x)) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}-\frac {2 \cos ^{\frac {3}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{3 e (a \cos (d+e x)+b+c \sin (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[d + e*x]^(3/2)*(c*Cos[d + e*x] - a*Sin[d + e*x])*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))/(3*e*(b
+ a*Cos[d + e*x] + c*Sin[d + e*x])) + (8*b*Cos[d + e*x]^(3/2)*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^
2 + c^2])/(b + Sqrt[a^2 + c^2])]*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))/(3*e*(b + a*Cos[d + e*x] + c*Sin
[d + e*x])*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])]) + (2*(a^2 - b^2 + c^2)*Cos[d + e
*x]^(3/2)*EllipticF[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[(b + a*Cos[d +
e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])]*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))/(3*e*(b + a*Cos[d
+ e*x] + c*Sin[d + e*x])^2)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3119

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*C
os[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])], Int[Sqrt[a/(a
+ Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]

Rule 3120

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> -Simp[((c*Cos[d
+ e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1))/(e*n), x] + Dist[1/n, Int[Simp[n*a^2 +
(n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x], x]*(a + b*Cos[d + e*x] + c*Sin
[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]

Rule 3127

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a +
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], Int[1/Sqrt[
a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; Free
Q[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]

Rule 3149

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.)
+ (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[B/b, Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]
, x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{a, b, c, d, e
, A, B, C}, x] && EqQ[B*c - b*C, 0] && NeQ[A*b - a*B, 0]

Rule 3163

Int[cos[(d_.) + (e_.)*(x_)]^(n_)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(n_),
x_Symbol] :> Dist[(Cos[d + e*x]^n*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^n)/(b + a*Cos[d + e*x] + c*Sin[d + e*
x])^n, Int[(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] &&  !IntegerQ[n]

Rubi steps

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

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Mathematica [F]  time = 150.81, size = 0, normalized size = 0.00 \[ \int \cos ^{\frac {3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (e x + d\right ) \sec \left (e x + d\right ) + c \cos \left (e x + d\right ) \tan \left (e x + d\right ) + a \cos \left (e x + d\right )\right )} \sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*cos(e*x + d)*sec(e*x + d) + c*cos(e*x + d)*tan(e*x + d) + a*cos(e*x + d))*sqrt(b*sec(e*x + d) + c*
tan(e*x + d) + a)*sqrt(cos(e*x + d)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {3}{2}} \cos \left (e x + d\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.29, size = 20776, normalized size = 56.00 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {3}{2}} \cos \left (e x + d\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (d+e\,x\right )}^{3/2}\,{\left (a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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