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

Optimal. Leaf size=170 \[ \frac{\sqrt{b} \left (15 a^2-20 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{8 d}+\frac{(a-b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{d}+\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac{b (7 a-4 b) \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)}}{8 d} \]

[Out]

((a - b)^(5/2)*ArcTan[(Sqrt[a - b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^2]])/d + (Sqrt[b]*(15*a^2 - 20*a*b +
8*b^2)*ArcTanh[(Sqrt[b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^2]])/(8*d) + ((7*a - 4*b)*b*Tan[c + d*x]*Sqrt[a
+ b*Tan[c + d*x]^2])/(8*d) + (b*Tan[c + d*x]*(a + b*Tan[c + d*x]^2)^(3/2))/(4*d)

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Rubi [A]  time = 0.178838, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3661, 416, 528, 523, 217, 206, 377, 203} \[ \frac{\sqrt{b} \left (15 a^2-20 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{8 d}+\frac{(a-b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{d}+\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac{b (7 a-4 b) \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)}}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^2)^(5/2),x]

[Out]

((a - b)^(5/2)*ArcTan[(Sqrt[a - b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^2]])/d + (Sqrt[b]*(15*a^2 - 20*a*b +
8*b^2)*ArcTanh[(Sqrt[b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^2]])/(8*d) + ((7*a - 4*b)*b*Tan[c + d*x]*Sqrt[a
+ b*Tan[c + d*x]^2])/(8*d) + (b*Tan[c + d*x]*(a + b*Tan[c + d*x]^2)^(3/2))/(4*d)

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])

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 528

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

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 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])

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 \left (a+b \tan ^2(c+d x)\right )^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{5/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2} \left (a (4 a-b)+(7 a-4 b) b x^2\right )}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 d}\\ &=\frac{(7 a-4 b) b \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)}}{8 d}+\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{a \left (8 a^2-9 a b+4 b^2\right )+b \left (15 a^2-20 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac{(7 a-4 b) b \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)}}{8 d}+\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac{(a-b)^3 \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (b \left (15 a^2-20 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac{(7 a-4 b) b \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)}}{8 d}+\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac{(a-b)^3 \operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{d}+\frac{\left (b \left (15 a^2-20 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{8 d}\\ &=\frac{(a-b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{d}+\frac{\sqrt{b} \left (15 a^2-20 a b+8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a+b \tan ^2(c+d x)}}\right )}{8 d}+\frac{(7 a-4 b) b \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)}}{8 d}+\frac{b \tan (c+d x) \left (a+b \tan ^2(c+d x)\right )^{3/2}}{4 d}\\ \end{align*}

Mathematica [C]  time = 1.24952, size = 259, normalized size = 1.52 \[ \frac{\sqrt{b} \left (15 a^2-20 a b+8 b^2\right ) \log \left (\sqrt{b} \sqrt{a+b \tan ^2(c+d x)}+b \tan (c+d x)\right )+b \tan (c+d x) \sqrt{a+b \tan ^2(c+d x)} \left (9 a+2 b \tan ^2(c+d x)-4 b\right )-4 i (a-b)^{5/2} \log \left (-\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \tan ^2(c+d x)}+a-i b \tan (c+d x)\right )}{(a-b)^{7/2} (\tan (c+d x)+i)}\right )+4 i (a-b)^{5/2} \log \left (\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \tan ^2(c+d x)}+a+i b \tan (c+d x)\right )}{(a-b)^{7/2} (\tan (c+d x)-i)}\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*(a - b)^(5/2)*Log[((-4*I)*(a - I*b*Tan[c + d*x] + Sqrt[a - b]*Sqrt[a + b*Tan[c + d*x]^2]))/((a - b)^(7
/2)*(I + Tan[c + d*x]))] + (4*I)*(a - b)^(5/2)*Log[((4*I)*(a + I*b*Tan[c + d*x] + Sqrt[a - b]*Sqrt[a + b*Tan[c
+ d*x]^2]))/((a - b)^(7/2)*(-I + Tan[c + d*x]))] + Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*Log[b*Tan[c + d*x] + Sqr
t[b]*Sqrt[a + b*Tan[c + d*x]^2]] + b*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]^2]*(9*a - 4*b + 2*b*Tan[c + d*x]^2))
/(8*d)

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Maple [B]  time = 0.059, size = 461, normalized size = 2.7 \begin{align*}{\frac{{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{4\,d}\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{9\,ab\tan \left ( dx+c \right ) }{8\,d}\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{15\,{a}^{2}}{8\,d}\sqrt{b}\ln \left ( \sqrt{b}\tan \left ( dx+c \right ) +\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}} \right ) }-{\frac{{b}^{2}\tan \left ( dx+c \right ) }{2\,d}\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,a}{2\,d}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{b}\tan \left ( dx+c \right ) +\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}} \right ) }+{\frac{1}{d}{b}^{{\frac{5}{2}}}\ln \left ( \sqrt{b}\tan \left ( dx+c \right ) +\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}} \right ) }-{\frac{b}{d \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( dx+c \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) }+3\,{\frac{a\sqrt{{b}^{4} \left ( a-b \right ) }}{d \left ( a-b \right ) }\arctan \left ({\frac{ \left ( a-b \right ){b}^{2}\tan \left ( dx+c \right ) }{\sqrt{{b}^{4} \left ( a-b \right ) }\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}} \right ) }-3\,{\frac{{a}^{2}\sqrt{{b}^{4} \left ( a-b \right ) }}{db \left ( a-b \right ) }\arctan \left ({\frac{ \left ( a-b \right ){b}^{2}\tan \left ( dx+c \right ) }{\sqrt{{b}^{4} \left ( a-b \right ) }\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}} \right ) }+{\frac{{a}^{3}}{d{b}^{2} \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( dx+c \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c)^2)^(5/2),x)

[Out]

1/4/d*b^2*tan(d*x+c)^3*(a+b*tan(d*x+c)^2)^(1/2)+9/8/d*b*a*tan(d*x+c)*(a+b*tan(d*x+c)^2)^(1/2)+15/8/d*b^(1/2)*a
^2*ln(b^(1/2)*tan(d*x+c)+(a+b*tan(d*x+c)^2)^(1/2))-1/2/d*b^2*tan(d*x+c)*(a+b*tan(d*x+c)^2)^(1/2)-5/2/d*b^(3/2)
*a*ln(b^(1/2)*tan(d*x+c)+(a+b*tan(d*x+c)^2)^(1/2))+1/d*b^(5/2)*ln(b^(1/2)*tan(d*x+c)+(a+b*tan(d*x+c)^2)^(1/2))
-1/d*b*(b^4*(a-b))^(1/2)/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))+3/d*a*(
b^4*(a-b))^(1/2)/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))-3/d*a^2/b*(b^4*
(a-b))^(1/2)/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))+1/d*a^3*(b^4*(a-b))
^(1/2)/b^2/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right )^{2} + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(d*x + c)^2 + a)^(5/2), x)

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Fricas [A]  time = 11.1738, size = 1763, normalized size = 10.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((15*a^2 - 20*a*b + 8*b^2)*sqrt(b)*log(2*b*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2 + a)*sqrt(b)*tan(d*x
+ c) + a) + 8*(a^2 - 2*a*b + b^2)*sqrt(-a + b)*log(-((a - 2*b)*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2 + a)*
sqrt(-a + b)*tan(d*x + c) - a)/(tan(d*x + c)^2 + 1)) + 2*(2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))
*sqrt(b*tan(d*x + c)^2 + a))/d, 1/16*(16*(a^2 - 2*a*b + b^2)*sqrt(a - b)*arctan(-sqrt(b*tan(d*x + c)^2 + a)/(s
qrt(a - b)*tan(d*x + c))) + (15*a^2 - 20*a*b + 8*b^2)*sqrt(b)*log(2*b*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2
+ a)*sqrt(b)*tan(d*x + c) + a) + 2*(2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))*sqrt(b*tan(d*x + c)^
2 + a))/d, -1/8*((15*a^2 - 20*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(d*x + c)^2 + a)*sqrt(-b)/(b*tan(d*x + c)
)) - 4*(a^2 - 2*a*b + b^2)*sqrt(-a + b)*log(-((a - 2*b)*tan(d*x + c)^2 + 2*sqrt(b*tan(d*x + c)^2 + a)*sqrt(-a
+ b)*tan(d*x + c) - a)/(tan(d*x + c)^2 + 1)) - (2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))*sqrt(b*ta
n(d*x + c)^2 + a))/d, 1/8*(8*(a^2 - 2*a*b + b^2)*sqrt(a - b)*arctan(-sqrt(b*tan(d*x + c)^2 + a)/(sqrt(a - b)*t
an(d*x + c))) - (15*a^2 - 20*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(d*x + c)^2 + a)*sqrt(-b)/(b*tan(d*x + c))
) + (2*b^2*tan(d*x + c)^3 + (9*a*b - 4*b^2)*tan(d*x + c))*sqrt(b*tan(d*x + c)^2 + a))/d]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c)**2)**(5/2),x)

[Out]

Integral((a + b*tan(c + d*x)**2)**(5/2), x)

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

[Out]

Timed out