3.272 \(\int (a+a \tan ^2(c+d x))^{5/2} \, dx\)
Optimal. Leaf size=98 \[ \frac{3 a^2 \tan (c+d x) \sqrt{a \sec ^2(c+d x)}}{8 d}+\frac{3 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec ^2(c+d x)}}\right )}{8 d}+\frac{a \tan (c+d x) \left (a \sec ^2(c+d x)\right )^{3/2}}{4 d} \]
[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)
________________________________________________________________________________________
Rubi [A] time = 0.048974, antiderivative size = 98, normalized size of antiderivative = 1.,
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 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
]
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 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 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])
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*}
Mathematica [A] time = 0.215845, 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)
________________________________________________________________________________________
Maple [A] time = 0.021, size = 90, normalized size = 0.9 \begin{align*}{\frac{a\tan \left ( dx+c \right ) }{4\,d} \left ( a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{2}\tan \left ( dx+c \right ) }{8\,d}\sqrt{a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3}{8\,d}{a}^{{\frac{5}{2}}}\ln \left ( \sqrt{a}\tan \left ( dx+c \right ) +\sqrt{a+a \left ( \tan \left ( dx+c \right ) \right ) ^{2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*tan(d*x+c)^2)^(5/2),x)
[Out]
1/4/d*a*tan(d*x+c)*(a+a*tan(d*x+c)^2)^(3/2)+3/8/d*a^2*tan(d*x+c)*(a+a*tan(d*x+c)^2)^(1/2)+3/8/d*a^(5/2)*ln(a^(
1/2)*tan(d*x+c)+(a+a*tan(d*x+c)^2)^(1/2))
________________________________________________________________________________________
Maxima [B] time = 2.87545, size = 2388, normalized size = 24.37 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*tan(d*x+c)^2)^(5/2),x, algorithm="maxima")
[Out]
1/16*(176*a^2*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 48*a^2*cos(d*x + c)*sin(2*d*x + 2*c) - 48*a^2*cos(2*d*x + 2*
c)*sin(d*x + c) - 12*a^2*sin(d*x + c) + 4*(3*a^2*sin(7*d*x + 7*c) + 11*a^2*sin(5*d*x + 5*c) - 11*a^2*sin(3*d*x
+ 3*c) - 3*a^2*sin(d*x + c))*cos(8*d*x + 8*c) - 24*(2*a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 2*a^2*s
in(2*d*x + 2*c))*cos(7*d*x + 7*c) + 16*(11*a^2*sin(5*d*x + 5*c) - 11*a^2*sin(3*d*x + 3*c) - 3*a^2*sin(d*x + c)
)*cos(6*d*x + 6*c) - 88*(3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*cos(5*d*x + 5*c) - 24*(11*a^2*sin(3*
d*x + 3*c) + 3*a^2*sin(d*x + c))*cos(4*d*x + 4*c) + 3*(a^2*cos(8*d*x + 8*c)^2 + 16*a^2*cos(6*d*x + 6*c)^2 + 36
*a^2*cos(4*d*x + 4*c)^2 + 16*a^2*cos(2*d*x + 2*c)^2 + a^2*sin(8*d*x + 8*c)^2 + 16*a^2*sin(6*d*x + 6*c)^2 + 36*
a^2*sin(4*d*x + 4*c)^2 + 48*a^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*a^2*sin(2*d*x + 2*c)^2 + 8*a^2*cos(2*d*
x + 2*c) + a^2 + 2*(4*a^2*cos(6*d*x + 6*c) + 6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos(2*d*x + 2*c) + a^2)*cos(8*d*x
+ 8*c) + 8*(6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos(2*d*x + 2*c) + a^2)*cos(6*d*x + 6*c) + 12*(4*a^2*cos(2*d*x + 2*
c) + a^2)*cos(4*d*x + 4*c) + 4*(2*a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*sin(
8*d*x + 8*c) + 16*(3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(cos(d*x + c)^2 + sin
(d*x + c)^2 + 2*sin(d*x + c) + 1) - 3*(a^2*cos(8*d*x + 8*c)^2 + 16*a^2*cos(6*d*x + 6*c)^2 + 36*a^2*cos(4*d*x +
4*c)^2 + 16*a^2*cos(2*d*x + 2*c)^2 + a^2*sin(8*d*x + 8*c)^2 + 16*a^2*sin(6*d*x + 6*c)^2 + 36*a^2*sin(4*d*x +
4*c)^2 + 48*a^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*a^2*sin(2*d*x + 2*c)^2 + 8*a^2*cos(2*d*x + 2*c) + a^2 +
2*(4*a^2*cos(6*d*x + 6*c) + 6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos(2*d*x + 2*c) + a^2)*cos(8*d*x + 8*c) + 8*(6*a^
2*cos(4*d*x + 4*c) + 4*a^2*cos(2*d*x + 2*c) + a^2)*cos(6*d*x + 6*c) + 12*(4*a^2*cos(2*d*x + 2*c) + a^2)*cos(4*
d*x + 4*c) + 4*(2*a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 1
6*(3*a^2*sin(4*d*x + 4*c) + 2*a^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*
sin(d*x + c) + 1) - 4*(3*a^2*cos(7*d*x + 7*c) + 11*a^2*cos(5*d*x + 5*c) - 11*a^2*cos(3*d*x + 3*c) - 3*a^2*cos(
d*x + c))*sin(8*d*x + 8*c) + 12*(4*a^2*cos(6*d*x + 6*c) + 6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos(2*d*x + 2*c) + a^
2)*sin(7*d*x + 7*c) - 16*(11*a^2*cos(5*d*x + 5*c) - 11*a^2*cos(3*d*x + 3*c) - 3*a^2*cos(d*x + c))*sin(6*d*x +
6*c) + 44*(6*a^2*cos(4*d*x + 4*c) + 4*a^2*cos(2*d*x + 2*c) + a^2)*sin(5*d*x + 5*c) + 24*(11*a^2*cos(3*d*x + 3*
c) + 3*a^2*cos(d*x + c))*sin(4*d*x + 4*c) - 44*(4*a^2*cos(2*d*x + 2*c) + a^2)*sin(3*d*x + 3*c))*sqrt(a)/((2*(4
*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*
cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) +
1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4
*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))
*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*
sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*d)
________________________________________________________________________________________
Fricas [A] time = 1.42479, size = 236, normalized size = 2.41 \begin{align*} \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} \end{align*}
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
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac{5}{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*tan(d*x+c)**2)**(5/2),x)
[Out]
Integral((a*tan(c + d*x)**2 + a)**(5/2), x)
________________________________________________________________________________________
Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
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