\(\int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [445]

Optimal result

Integrand size = 35, antiderivative size = 241 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]

[Out]

(I*a-b)^(5/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+b^(3/2)*(2*A*b+5*B*a)*ar
ctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-(I*a+b)^(5/2)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c
)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+b*(2*A*a+B*b)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d-2*a*A*(a+b*tan(d*x+c
))^(3/2)/d/tan(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 2.97 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3686, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(-b+i a)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^{3/2} (5 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(b+i a)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]

[In]

Int[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(3/2),x]

[Out]

((I*a - b)^(5/2)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b^(3/2)*(
2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((I*a + b)^(5/2)*(A - I*B)*
ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b*(2*a*A + b*B)*Sqrt[Tan[c + d*x]]*
Sqrt[a + b*Tan[c + d*x]])/d - (2*a*A*(a + b*Tan[c + d*x])^(3/2))/(d*Sqrt[Tan[c + d*x]])

Rule 65

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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

Rule 223

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 3686

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3736

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {1}{2} a (4 A b+a B)-\frac {1}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {1}{2} b (2 a A+b B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {\frac {1}{4} a \left (6 a A b+2 a^2 B-b^2 B\right )-\frac {1}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)+\frac {1}{4} b^2 (2 A b+5 a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {\frac {1}{4} a \left (6 a A b+2 a^2 B-b^2 B\right )+\frac {1}{2} \left (-a^3 A+3 a A b^2+3 a^2 b B-b^3 B\right ) x+\frac {1}{4} b^2 (2 A b+5 a B) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \left (\frac {b^2 (2 A b+5 a B)}{4 \sqrt {x} \sqrt {a+b x}}+\frac {3 a^2 A b-A b^3+a^3 B-3 a b^2 B-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x}{2 \sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {3 a^2 A b-A b^3+a^3 B-3 a b^2 B-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b^2 (2 A b+5 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {\text {Subst}\left (\int \left (\frac {a^3 A-3 a A b^2-3 a^2 b B+b^3 B+i \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-a^3 A+3 a A b^2+3 a^2 b B-b^3 B+i \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b^2 (2 A b+5 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left ((a-i b)^3 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left ((a+i b)^3 (A+i B)\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (b^2 (2 A b+5 a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left ((a-i b)^3 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left ((a+i b)^3 (A+i B)\right ) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b (2 a A+b B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.71 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(-1)^{3/4} (-a+i b)^{5/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\sqrt [4]{-1} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {(a-i b)^2 (A-i B) \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}-\frac {(a+i b)^2 (A+i B) \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}+\frac {b B (a+b \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}}+\frac {b (2 A b+5 a B) \sqrt {a+b \tan (c+d x)} \left (\frac {\sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {\tan (c+d x)}}\right )}{\sqrt {1+\frac {b \tan (c+d x)}{a}}}}{d} \]

[In]

Integrate[((a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(3/2),x]

[Out]

((-1)^(3/4)*(-a + I*b)^(5/2)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]] - (-1)^(1/4)*(a + I*b)^(5/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a +
b*Tan[c + d*x]]] - ((a - I*b)^2*(A - I*B)*Sqrt[a + b*Tan[c + d*x]])/Sqrt[Tan[c + d*x]] - ((a + I*b)^2*(A + I*B
)*Sqrt[a + b*Tan[c + d*x]])/Sqrt[Tan[c + d*x]] + (b*B*(a + b*Tan[c + d*x])^(3/2))/Sqrt[Tan[c + d*x]] + (b*(2*A
*b + 5*a*B)*Sqrt[a + b*Tan[c + d*x]]*((Sqrt[b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[a] - Sqrt[1
+ (b*Tan[c + d*x])/a]/Sqrt[Tan[c + d*x]]))/Sqrt[1 + (b*Tan[c + d*x])/a])/d

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.08 (sec) , antiderivative size = 2653772, normalized size of antiderivative = 11011.50

\[\text {output too large to display}\]

[In]

int((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17614 vs. \(2 (197) = 394\).

Time = 7.59 (sec) , antiderivative size = 35230, normalized size of antiderivative = 146.18 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(3/2),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(5/2)/tan(c + d*x)**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/tan(c + d*x)^(3/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2))/tan(c + d*x)^(3/2), x)