∫ (a + a tan2(c + dx))5∕2 dx

3.272 \(\int (a+a \tan ^2(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=98 \[ \frac {3 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec ^2(c+d x)}}\right )}{8 d}+\frac {3 a^2 \tan (c+d x) \sqrt {a \sec ^2(c+d x)}}{8 d}+\frac {a \tan (c+d x) \left (a \sec ^2(c+d x)\right )^{3/2}}{4 d} \]

[Out]

3/8*a^(5/2)*arctanh(a^(1/2)*tan(d*x+c)/(a*sec(d*x+c)^2)^(1/2))/d+1/4*a*(a*sec(d*x+c)^2)^(3/2)*tan(d*x+c)/d+3/8

*a^2*(a*sec(d*x+c)^2)^(1/2)*tan(d*x+c)/d

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Rubi [A] time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3657, 4122, 195, 217, 206} \[ \frac {3 a^2 \tan (c+d x) \sqrt {a \sec ^2(c+d x)}}{8 d}+\frac {3 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec ^2(c+d x)}}\right )}{8 d}+\frac {a \tan (c+d x) \left (a \sec ^2(c+d x)\right )^{3/2}}{4 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[(a + a*Tan[c + d*x]^2)^(5/2),x]

[Out]

(3*a^(5/2)*ArcTanh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a*Sec[c + d*x]^2]])/(8*d) + (3*a^2*Sqrt[a*Sec[c + d*x]^2]*Tan[c

+ d*x])/(8*d) + (a*(a*Sec[c + d*x]^2)^(3/2)*Tan[c + d*x])/(4*d)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),

Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2

] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]

/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,

b}, x] && !GtQ[a, 0]

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]

, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)

/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] && !IntegerQ[p

]

Rubi steps

\begin {align*} \int \left (a+a \tan ^2(c+d x)\right )^{5/2} \, dx &=\int \left (a \sec ^2(c+d x)\right )^{5/2} \, dx\\ &=\frac {a \operatorname {Subst}\left (\int \left (a+a x^2\right )^{3/2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {a \left (a \sec ^2(c+d x)\right )^{3/2} \tan (c+d x)}{4 d}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \sqrt {a+a x^2} \, dx,x,\tan (c+d x)\right )}{4 d}\\ &=\frac {3 a^2 \sqrt {a \sec ^2(c+d x)} \tan (c+d x)}{8 d}+\frac {a \left (a \sec ^2(c+d x)\right )^{3/2} \tan (c+d x)}{4 d}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+a x^2}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {3 a^2 \sqrt {a \sec ^2(c+d x)} \tan (c+d x)}{8 d}+\frac {a \left (a \sec ^2(c+d x)\right )^{3/2} \tan (c+d x)}{4 d}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\tan (c+d x)}{\sqrt {a \sec ^2(c+d x)}}\right )}{8 d}\\ &=\frac {3 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec ^2(c+d x)}}\right )}{8 d}+\frac {3 a^2 \sqrt {a \sec ^2(c+d x)} \tan (c+d x)}{8 d}+\frac {a \left (a \sec ^2(c+d x)\right )^{3/2} \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A] time = 0.22, size = 65, normalized size = 0.66 \[ \frac {a^2 \cos (c+d x) \sqrt {a \sec ^2(c+d x)} \left (3 \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (2 \sec ^2(c+d x)+3\right )\right )}{8 d} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + a*Tan[c + d*x]^2)^(5/2),x]

[Out]

(a^2*Cos[c + d*x]*Sqrt[a*Sec[c + d*x]^2]*(3*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(3 + 2*Sec[c + d*x]^2)*Tan[c

+ d*x]))/(8*d)

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fricas [A] time = 0.40, size = 91, normalized size = 0.93 \[ \frac {3 \, a^{\frac {5}{2}} \log \left (2 \, a \tan \left (d x + c\right )^{2} + 2 \, \sqrt {a \tan \left (d x + c\right )^{2} + a} \sqrt {a} \tan \left (d x + c\right ) + a\right ) + 2 \, {\left (2 \, a^{2} \tan \left (d x + c\right )^{3} + 5 \, a^{2} \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{16 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(d*x+c)^2)^(5/2),x, algorithm="fricas")

[Out]

1/16*(3*a^(5/2)*log(2*a*tan(d*x + c)^2 + 2*sqrt(a*tan(d*x + c)^2 + a)*sqrt(a)*tan(d*x + c) + a) + 2*(2*a^2*tan

(d*x + c)^3 + 5*a^2*tan(d*x + c))*sqrt(a*tan(d*x + c)^2 + a))/d

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(d*x+c)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/