3.112 \(\int (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=125 \[ \frac {b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f}+\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} (3 a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f} \]

[Out]

(a-b)^(3/2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f+1/2*(3*a-2*b)*arctanh(b^(1/2)*tan(f*x+e)
/(a+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f+1/2*b*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3661, 416, 523, 217, 206, 377, 203} \[ \frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f}+\frac {\sqrt {b} (3 a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f + ((3*a - 2*b)*Sqrt[b]*ArcTanh
[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) + (b*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*f)

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])

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 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 416

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [C]  time = 1.47, size = 233, normalized size = 1.86 \[ \frac {b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}-i (a-b)^{3/2} \log \left (-\frac {4 i \left (\sqrt {a-b} \sqrt {a+b \tan ^2(e+f x)}+a-i b \tan (e+f x)\right )}{(a-b)^{5/2} (\tan (e+f x)+i)}\right )+i (a-b)^{3/2} \log \left (\frac {4 i \left (\sqrt {a-b} \sqrt {a+b \tan ^2(e+f x)}+a+i b \tan (e+f x)\right )}{(a-b)^{5/2} (\tan (e+f x)-i)}\right )+\sqrt {b} (3 a-2 b) \log \left (\sqrt {b} \sqrt {a+b \tan ^2(e+f x)}+b \tan (e+f x)\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((-I)*(a - b)^(3/2)*Log[((-4*I)*(a - I*b*Tan[e + f*x] + Sqrt[a - b]*Sqrt[a + b*Tan[e + f*x]^2]))/((a - b)^(5/2
)*(I + Tan[e + f*x]))] + I*(a - b)^(3/2)*Log[((4*I)*(a + I*b*Tan[e + f*x] + Sqrt[a - b]*Sqrt[a + b*Tan[e + f*x
]^2]))/((a - b)^(5/2)*(-I + Tan[e + f*x]))] + (3*a - 2*b)*Sqrt[b]*Log[b*Tan[e + f*x] + Sqrt[b]*Sqrt[a + b*Tan[
e + f*x]^2]] + b*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*f)

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fricas [A]  time = 0.88, size = 537, normalized size = 4.30 \[ \left [-\frac {{\left (3 \, a - 2 \, b\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, {\left (a - b\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, f}, -\frac {{\left (3 \, a - 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) - {\left (-a + b\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, f}, \frac {4 \, {\left (a - b\right )}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - {\left (3 \, a - 2 \, b\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, f}, \frac {2 \, {\left (a - b\right )}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - {\left (3 \, a - 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) + \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((3*a - 2*b)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) + 2
*(a - b)*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e)
- a)/(tan(f*x + e)^2 + 1)) - 2*sqrt(b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/f, -1/2*((3*a - 2*b)*sqrt(-b)*arctan
(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) - (-a + b)^(3/2)*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqr
t(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - sqrt(b*tan(f*x + e)^2 + a)*b*ta
n(f*x + e))/f, 1/4*(4*(a - b)^(3/2)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - (3*a - 2*
b)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) + 2*sqrt(b*tan(f*x
+ e)^2 + a)*b*tan(f*x + e))/f, 1/2*(2*(a - b)^(3/2)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x +
e))) - (3*a - 2*b)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + sqrt(b*tan(f*x + e)
^2 + a)*b*tan(f*x + e))/f]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2), x)

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maple [B]  time = 0.26, size = 297, normalized size = 2.38 \[ \frac {b \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 f}-\frac {b^{\frac {3}{2}} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \left (a -b \right )}-\frac {2 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f b \left (a -b \right )}+\frac {a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \,b^{2} \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

1/2*b*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f+3/2/f*b^(1/2)*a*ln(tan(f*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))-1/
f*b^(3/2)*ln(tan(f*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))+1/f*(b^4*(a-b))^(1/2)/(a-b)*arctan((a-b)*b^2/(b^4*(a
-b))^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))-2/f*a/b*(b^4*(a-b))^(1/2)/(a-b)*arctan((a-b)*b^2/(b^4*(a-b))^(
1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))+1/f*a^2*(b^4*(a-b))^(1/2)/b^2/(a-b)*arctan((a-b)*b^2/(b^4*(a-b))^(1/
2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int((a + b*tan(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x)**2)**(3/2), x)

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