3.625 \(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=231 \[ \frac {\sqrt {b} \left (15 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {(-b+i a)^{5/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

[Out]

-(I*a-b)^(5/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+(I*a+b)^(5/2)*arctanh((I*a+b)^(
1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+1/4*(15*a^2-8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d
*x+c))^(1/2))*b^(1/2)/d+9/4*a*b*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d+1/2*b^2*(a+b*tan(d*x+c))^(1/2)*tan(d
*x+c)^(3/2)/d

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Rubi [A]  time = 1.89, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3566, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac {\sqrt {b} \left (15 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {(-b+i a)^{5/2} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {(b+i a)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (Sqrt[b]*(15*a^2
- 8*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(4*d) + ((I*a + b)^(5/2)*ArcTanh[(Sqr
t[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (9*a*b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x
]])/(4*d) + (b^2*Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(2*d)

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 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 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 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]

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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

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]

Rubi steps

\begin {align*} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx &=\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{2} \int \frac {\sqrt {\tan (c+d x)} \left (\frac {1}{2} a \left (4 a^2-3 b^2\right )+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {9}{2} a b^2 \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\int \frac {-\frac {9}{4} a^2 b^2+2 a b \left (a^2-3 b^2\right ) \tan (c+d x)+\frac {1}{4} b^2 \left (15 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 b}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {9}{4} a^2 b^2+2 a b \left (a^2-3 b^2\right ) x+\frac {1}{4} b^2 \left (15 a^2-8 b^2\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {b^2 \left (15 a^2-8 b^2\right )}{4 \sqrt {x} \sqrt {a+b x}}-\frac {2 \left (3 a^2 b^2-b^4-a b \left (a^2-3 b^2\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {3 a^2 b^2-b^4-a b \left (a^2-3 b^2\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (b \left (15 a^2-8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {\operatorname {Subst}\left (\int \left (\frac {a b \left (a^2-3 b^2\right )+i \left (3 a^2 b^2-b^4\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-a b \left (a^2-3 b^2\right )+i \left (3 a^2 b^2-b^4\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (b \left (15 a^2-8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {(a-i b)^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {(a+i b)^3 \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 \left (15 a^2-8 b^2\right )\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 )}{4 d}\\ &=\frac {\sqrt {b} \left (15 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {(a-i b)^3 \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 {(a+i b)^3 \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} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\\ \end {align*}

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Mathematica [A]  time = 2.04, size = 264, normalized size = 1.14 \[ \frac {\frac {\sqrt {a} \sqrt {b} \left (15 a^2-8 b^2\right ) \sqrt {\frac {b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a+b \tan (c+d x)}}+2 b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+4 \sqrt [4]{-1} (-a+i b)^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 \sqrt [4]{-1} (a+i b)^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]
+ 4*(-1)^(1/4)*(a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]
+ 9*a*b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*b^2*Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]] + (Sq
rt[a]*Sqrt[b]*(15*a^2 - 8*b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqr
t[a + b*Tan[c + d*x]])/(4*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.78, size = 1345303, normalized size = 5823.82 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\tan \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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