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

Optimal. Leaf size=241 \[ \frac{b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b (2 a A+b B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt{\tan (c+d x)}} \]

[Out]

((I*a - b)^(5/2)*(A + I*B)*ArcTan[(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]])

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Rubi [A]  time = 2.3508, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3605, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b (2 a A+b B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+b \tan (c+d x))^{3/2}}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[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 3605

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 3647

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 3655

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 6725

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]

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 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 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 205

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

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{(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{3}{2}}(c+d x)} \, dx &=-\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 \operatorname{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 \operatorname{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{\operatorname{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 ) \operatorname{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{\operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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) \tanh ^{-1}\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 ) \operatorname{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 ) \operatorname{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) \tan ^{-1}\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) \tanh ^{-1}\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) \tanh ^{-1}\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 [C]  time = 40.969, size = 209298, normalized size = 868.46 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.92, size = 2652458, normalized size = 11006.1 \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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x)

[Out]

result too large to display

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

[Out]

Timed out

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

[Out]

Timed out