3.554 \(\int \frac{1}{(a+b \tan (c+d x))^{7/2}} \, dx\)
Optimal. Leaf size=194 \[ -\frac{2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 \sqrt{a+b \tan (c+d x)}}-\frac{4 a b}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}-\frac{2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{7/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{7/2}} \]
[Out]
((-I)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(7/2)*d) + (I*ArcTanh[Sqrt[a + b*Tan[c + d*x
]]/Sqrt[a + I*b]])/((a + I*b)^(7/2)*d) - (2*b)/(5*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(5/2)) - (4*a*b)/(3*(a^2
+ b^2)^2*d*(a + b*Tan[c + d*x])^(3/2)) - (2*b*(3*a^2 - b^2))/((a^2 + b^2)^3*d*Sqrt[a + b*Tan[c + d*x]])
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Rubi [A] time = 0.388807, antiderivative size = 194, normalized size of antiderivative = 1.,
number of steps used = 10, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used
= {3483, 3529, 3539, 3537, 63, 208} \[ -\frac{2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 \sqrt{a+b \tan (c+d x)}}-\frac{4 a b}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}-\frac{2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (a-i b)^{7/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (a+i b)^{7/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[c + d*x])^(-7/2),x]
[Out]
((-I)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((a - I*b)^(7/2)*d) + (I*ArcTanh[Sqrt[a + b*Tan[c + d*x
]]/Sqrt[a + I*b]])/((a + I*b)^(7/2)*d) - (2*b)/(5*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(5/2)) - (4*a*b)/(3*(a^2
+ b^2)^2*d*(a + b*Tan[c + d*x])^(3/2)) - (2*b*(3*a^2 - b^2))/((a^2 + b^2)^3*d*Sqrt[a + b*Tan[c + d*x]])
Rule 3483
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
Rule 3529
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]
Rule 3539
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Rule 3537
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{(a+b \tan (c+d x))^{7/2}} \, dx &=-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}+\frac{\int \frac{a-b \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx}{a^2+b^2}\\ &=-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac{4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}+\frac{\int \frac{a^2-b^2-2 a b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac{4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3 d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{\left (a^2+b^2\right )^3}\\ &=-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac{4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3 d \sqrt{a+b \tan (c+d x)}}+\frac{\int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^3}+\frac{\int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^3}\\ &=-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac{4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3 d \sqrt{a+b \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 (i a-b)^3 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 (i a+b)^3 d}\\ &=-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac{4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3 d \sqrt{a+b \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{(a-i b)^3 b d}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{(a+i b)^3 b d}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{(a-i b)^{7/2} d}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{(a+i b)^{7/2} d}-\frac{2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac{4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}-\frac{2 b \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 0.210104, size = 108, normalized size = 0.56 \[ \frac{i (a+i b) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{a+b \tan (c+d x)}{a-i b}\right )+(-b-i a) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{a+b \tan (c+d x)}{a+i b}\right )}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[c + d*x])^(-7/2),x]
[Out]
(I*(a + I*b)*Hypergeometric2F1[-5/2, 1, -3/2, (a + b*Tan[c + d*x])/(a - I*b)] + ((-I)*a - b)*Hypergeometric2F1
[-5/2, 1, -3/2, (a + b*Tan[c + d*x])/(a + I*b)])/(5*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(5/2))
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Maple [B] time = 0.055, size = 3082, normalized size = 15.9 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*tan(d*x+c))^(7/2),x)
[Out]
1/d/b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(
1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^5-3/d*b^3/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2
+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/d/b/(a^2+b^2)^(9/2)/(2*(a^
2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a
)^(1/2))*a^7+5/d*b^3/(a^2+b^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+
b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+7/d*b^5/(a^2+b^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*
arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+5/4/d*b^3/(a^
2+b^2)^(9/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b
^2)^(1/2)+2*a)^(1/2)*a^2-1/d/b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+
(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^5+3/d*b^3/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2
*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/d
/b/(a^2+b^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2
))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^7-5/d*b^3/(a^2+b^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan
(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3-7/d*b^5/(a^2+b^2)^(9/2)/(2*(a
^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*
a)^(1/2))*a+1/4/d/b/(a^2+b^2)^(9/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^
2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^6-5/4/d*b/(a^2+b^2)^(9/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)
^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-2/d*b/(a^2+b^2)^(7/2)/(2*(
a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2
*a)^(1/2))*a^3-3/d*b/(a^2+b^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+
b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^5-1/2/d*b/(a^2+b^2)^4*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))
^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+1/2/d*b/(a^2+b^2)^4*ln
((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^
(1/2)*a^3+3/d*b/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x
+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^4-3/d*b/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*
tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^4-2/5*b/(a^2+b^2)/d/(a+b*tan
(d*x+c))^(5/2)+5/4/d*b/(a^2+b^2)^(9/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+
(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-1/d*b^5/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((
2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d*b^5/(a^2+b^2)^(9/2
)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2
*a)^(1/2)-1/4/d*b^5/(a^2+b^2)^(9/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^
2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-6/d*b/(a^2+b^2)^3/(a+b*tan(d*x+c))^(1/2)*a^2+1/d*b^5/(a^2+b^2)^4/(
2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2
)-2*a)^(1/2))+2/d*b^3/(a^2+b^2)^3/(a+b*tan(d*x+c))^(1/2)+2/d*b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*a
rctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+3/d*b/(a^2+b
^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^
2+b^2)^(1/2)-2*a)^(1/2))*a^5-3/4/d*b^3/(a^2+b^2)^4*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)
+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-2/d*b^3/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)
*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2-1/4/d/b/(a
^2+b^2)^4*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)
^(1/2)+2*a)^(1/2)*a^5+3/4/d*b^3/(a^2+b^2)^4*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+
c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+2/d*b^3/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan
(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2+1/4/d/b/(a^2+b^2)
^4*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+
2*a)^(1/2)*a^5-5/4/d*b^3/(a^2+b^2)^(9/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-
a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d/b/(a^2+b^2)^(9/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c)
)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^6-4/3*a*b/(a^2+b^2)^2/d
/(a+b*tan(d*x+c))^(3/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [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(1/(a+b*tan(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{7}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c))**(7/2),x)
[Out]
Integral((a + b*tan(c + d*x))**(-7/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*tan(d*x+c))^(7/2),x, algorithm="giac")
[Out]
integrate((b*tan(d*x + c) + a)^(-7/2), x)