3.622 \(\int \frac{(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=266 \[ \frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}-\frac{(-b+i a)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{(b+i a)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

[Out]

-(((I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) - ((I*a + b)^(3/2)*
ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*a*Sqrt[a + b*Tan[c + d*x]])/(7*d*
Tan[c + d*x]^(7/2)) - (16*b*Sqrt[a + b*Tan[c + d*x]])/(35*d*Tan[c + d*x]^(5/2)) + (2*(35*a^2 - 3*b^2)*Sqrt[a +
b*Tan[c + d*x]])/(105*a*d*Tan[c + d*x]^(3/2)) + (4*b*(70*a^2 + 3*b^2)*Sqrt[a + b*Tan[c + d*x]])/(105*a^2*d*Sq
rt[Tan[c + d*x]])

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Rubi [A]  time = 1.13896, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3567, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}-\frac{(-b+i a)^{3/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{(b+i a)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(9/2),x]

[Out]

-(((I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) - ((I*a + b)^(3/2)*
ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*a*Sqrt[a + b*Tan[c + d*x]])/(7*d*
Tan[c + d*x]^(7/2)) - (16*b*Sqrt[a + b*Tan[c + d*x]])/(35*d*Tan[c + d*x]^(5/2)) + (2*(35*a^2 - 3*b^2)*Sqrt[a +
b*Tan[c + d*x]])/(105*a*d*Tan[c + d*x]^(3/2)) + (4*b*(70*a^2 + 3*b^2)*Sqrt[a + b*Tan[c + d*x]])/(105*a^2*d*Sq
rt[Tan[c + d*x]])

Rule 3567

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d
^2, 0] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3649

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*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3616

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

Rule 3615

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

Rule 93

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 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])

Rubi steps

\begin{align*} \int \frac{(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac{9}{2}}(c+d x)} \, dx &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2}{7} \int \frac{-4 a b+\frac{7}{2} \left (a^2-b^2\right ) \tan (c+d x)+3 a b \tan ^2(c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{-\frac{1}{4} a \left (35 a^2-3 b^2\right )-\frac{35}{2} a^2 b \tan (c+d x)-8 a b^2 \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{35 a}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 \int \frac{\frac{1}{4} a b \left (70 a^2+3 b^2\right )-\frac{105}{8} a^2 \left (a^2-b^2\right ) \tan (c+d x)-\frac{1}{4} a b \left (35 a^2-3 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{105 a^2}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}+\frac{16 \int \frac{\frac{105}{16} a^3 \left (a^2-b^2\right )+\frac{105}{8} a^4 b \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}+\frac{1}{2} (a-i b)^2 \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx+\frac{1}{2} (a+i b)^2 \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}+\frac{(a-i b)^2 \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(a+i b)^2 \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}+\frac{(a-i b)^2 \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(a+i b)^2 \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac{(i a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(i a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a \sqrt{a+b \tan (c+d x)}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{16 b \sqrt{a+b \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (35 a^2-3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 b \left (70 a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{105 a^2 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.82334, size = 222, normalized size = 0.83 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (a \left (35 a^2-3 b^2\right ) \tan ^2(c+d x)+2 b \left (70 a^2+3 b^2\right ) \tan ^3(c+d x)-24 a^2 b \tan (c+d x)-15 a^3\right )}{a^2 \tan ^{\frac{7}{2}}(c+d x)}-105 (-1)^{3/4} (-a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-105 (-1)^{3/4} (a-i b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(9/2),x]

[Out]

(-105*(-1)^(3/4)*(-a - I*b)^(3/2)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*
x]]] - 105*(-1)^(3/4)*(a - I*b)^(3/2)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c +
d*x]]] + (2*Sqrt[a + b*Tan[c + d*x]]*(-15*a^3 - 24*a^2*b*Tan[c + d*x] + a*(35*a^2 - 3*b^2)*Tan[c + d*x]^2 + 2
*b*(70*a^2 + 3*b^2)*Tan[c + d*x]^3))/(a^2*Tan[c + d*x]^(7/2)))/(105*d)

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Maple [B]  time = 0.342, size = 1346975, normalized size = 5063.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\tan \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(9/2), x)

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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((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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))**(3/2)/tan(d*x+c)**(9/2),x)

[Out]

Timed out

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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))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Timed out