\(\int \frac {(a+b \tan (c+d x))^{5/2} (\frac {3 b B}{2 a}+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [451]

Optimal result

Integrand size = 43, antiderivative size = 253 \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (2 a-3 i b) B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a d}+\frac {2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(2 a+3 i b) (i a+b)^{5/2} B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a d}-\frac {2 \left (a^2+3 b^2\right ) B \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 3.34 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3686, 3726, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 B \left (a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {B (2 a-3 i b) (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a d}+\frac {2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {B (2 a+3 i b) (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 a d}-\frac {b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

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

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 3726

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] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((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)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

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

Mathematica [A] (verified)

Time = 4.86 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {B \cos (c+d x) (3 b+2 a \tan (c+d x)) \left (4 \sqrt {a} b^{5/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}-\sqrt {1+\frac {b \tan (c+d x)}{a}} \left (\sqrt [4]{-1} (-a+i b)^{5/2} (2 a+3 i b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {3}{2}}(c+d x)+\sqrt [4]{-1} (a+i b)^{5/2} (2 a-3 i b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \tan ^{\frac {3}{2}}(c+d x)+2 a \sqrt {a+b \tan (c+d x)} \left (a b+\left (2 a^2+7 b^2\right ) \tan (c+d x)\right )\right )\right )}{2 a d (3 b \cos (c+d x)+2 a \sin (c+d x)) \tan ^{\frac {3}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]

[In]

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

[Out]

(B*Cos[c + d*x]*(3*b + 2*a*Tan[c + d*x])*(4*Sqrt[a]*b^(5/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Tan[
c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]] - Sqrt[1 + (b*Tan[c + d*x])/a]*((-1)^(1/4)*(-a + I*b)^(5/2)*(2*a + (3*
I)*b)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(3/2) + (-1
)^(1/4)*(a + I*b)^(5/2)*(2*a - (3*I)*b)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]]*Tan[c + d*x]^(3/2) + 2*a*Sqrt[a + b*Tan[c + d*x]]*(a*b + (2*a^2 + 7*b^2)*Tan[c + d*x]))))/(2*a*d*(3*b
*Cos[c + d*x] + 2*a*Sin[c + d*x])*Tan[c + d*x]^(3/2)*Sqrt[1 + (b*Tan[c + d*x])/a])

Maple [B] (warning: unable to verify)

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

Time = 5.86 (sec) , antiderivative size = 1491744, normalized size of antiderivative = 5896.22

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

[In]

int((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),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. 8171 vs. \(2 (205) = 410\).

Time = 2.54 (sec) , antiderivative size = 16341, normalized size of antiderivative = 64.59 \[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {B \left (\int \frac {2 a^{3} \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {3 b^{3} \sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \frac {6 a b^{2} \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int 2 a b^{2} \sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\tan {\left (c + d x \right )}}\, dx + \int \frac {3 a^{2} b \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4 a^{2} b \sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx\right )}{2 a} \]

[In]

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

[Out]

B*(Integral(2*a**3*sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(3/2), x) + Integral(3*b**3*sqrt(a + b*tan(c + d*x))
/sqrt(tan(c + d*x)), x) + Integral(6*a*b**2*sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(3/2), x) + Integral(2*a*b*
*2*sqrt(a + b*tan(c + d*x))*sqrt(tan(c + d*x)), x) + Integral(3*a**2*b*sqrt(a + b*tan(c + d*x))/tan(c + d*x)**
(5/2), x) + Integral(4*a**2*b*sqrt(a + b*tan(c + d*x))/sqrt(tan(c + d*x)), x))/(2*a)

Maxima [F]

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

[In]

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

[Out]

1/2*integrate((2*B*tan(d*x + c) + 3*B*b/a)*(b*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(5/2), x)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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