3.5.0 ∫ tan5 _2 (c+dx)(A+B tan(c+dx)) (a+b tan(c+dx))2 dx [404]

Integrand size = 33, antiderivative size = 436 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {a^{3/2} \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]

Rubi [A] (veriﬁed)

Time = 1.77 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3686, 3728, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2 (A-B)+2 a b (A+B)-b^2 (A-B)\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\left (-3 a^2 B+a A b-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 d \left (a^2+b^2\right )}+\frac {\left (-\left (a^2 (A+B)\right )+2 a b (A-B)+b^2 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (-\left (a^2 (A+B)\right )+2 a b (A-B)+b^2 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {a^{3/2} \left (-3 a^3 B+a^2 A b-7 a b^2 B+5 A b^3\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} d \left (a^2+b^2\right )^2} \]

Rubi steps \begin{align*} \text {integral}& = \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\frac {3}{2} a (A b-a B)+b (A b-a B) \tan (c+d x)-\frac {1}{2} \left (a A b-3 a^2 B-2 b^2 B\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )} \\ & = -\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {2 \int \frac {\frac {1}{4} a \left (a A b-3 a^2 B-2 b^2 B\right )-\frac {1}{2} b^2 (a A+b B) \tan (c+d x)+\frac {1}{4} \left (a^2 A b+2 A b^3-3 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{b^2 \left (a^2+b^2\right )} \\ & = -\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {2 \int \frac {-\frac {1}{2} b^2 \left (2 a A b-a^2 B+b^2 B\right )-\frac {1}{2} b^2 \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{b^2 \left (a^2+b^2\right )^2}+\frac {\left (a^2 \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 b^2 \left (a^2+b^2\right )^2} \\ & = -\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {4 \text {Subst}\left (\int \frac {-\frac {1}{2} b^2 \left (2 a A b-a^2 B+b^2 B\right )-\frac {1}{2} b^2 \left (a^2 A-A b^2+2 a b B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^2 \left (a^2+b^2\right )^2 d}+\frac {\left (a^2 \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{2 b^2 \left (a^2+b^2\right )^2 d} \\ & = -\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (a^2 \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b^2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d} \\ & = \frac {a^{3/2} \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}-\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d} \\ & = \frac {a^{3/2} \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d} \\ & = \frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {a^{3/2} \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}+\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [C] (veriﬁed)

Time = 2.49 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.63 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \left ((-A b+3 a B) \sqrt {\tan (c+d x)}+\frac {\left (a^2 A b+2 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right )}+b B \tan ^{\frac {3}{2}}(c+d x)-\frac {\left (\sqrt [4]{-1} (a+i b)^2 b^{5/2} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+a^{3/2} \left (-a^2 A b-5 A b^3+3 a^3 B+7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+(-1)^{3/4} b^{5/2} (i a+b)^2 (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right ) (a+b \tan (c+d x))}{2 \sqrt {b} \left (a^2+b^2\right )^2}\right )}{b^2 d (a+b \tan (c+d x))} \]

Maple [A] (veriﬁed)

Time = 0.08 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.81


method	result	size

derivativedivides	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{b^{2}}+\frac {2 a^{2} \left (\frac {\left (-\frac {1}{2} A \,a^{2} b -\frac {1}{2} A \,b^{3}+\frac {1}{2} B \,a^{3}+\frac {1}{2} B a \,b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}+\frac {\left (A \,a^{2} b +5 A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (-2 A a b +B \,a^{2}-B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A \,a^{2}+A \,b^{2}-2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(351\)
default	\(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{b^{2}}+\frac {2 a^{2} \left (\frac {\left (-\frac {1}{2} A \,a^{2} b -\frac {1}{2} A \,b^{3}+\frac {1}{2} B \,a^{3}+\frac {1}{2} B a \,b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{a +b \tan \left (d x +c \right )}+\frac {\left (A \,a^{2} b +5 A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (-2 A a b +B \,a^{2}-B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A \,a^{2}+A \,b^{2}-2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(351\)

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5942 vs. \(2 (397) = 794\).

Time = 36.20 (sec) , antiderivative size = 11910, normalized size of antiderivative = 27.32 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]

Sympy [F]

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

Maxima [A] (veriﬁcation not implemented)

Time = 0.32 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.86 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{5} - A a^{4} b + 7 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a b}} - \frac {4 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {\tan \left (d x + c\right )}}{a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )} + \frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} + 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} + 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left ({\left (A + B\right )} a^{2} - 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left ({\left (A + B\right )} a^{2} - 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {8 \, B \sqrt {\tan \left (d x + c\right )}}{b^{2}}}{4 \, d} \]

Giac [F(-1)]

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

Mupad [B] (veriﬁcation not implemented)

Time = 41.99 (sec) , antiderivative size = 18313, normalized size of antiderivative = 42.00 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]

\(\int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\) [404]

Optimal result