Optimal. Leaf size=260 \[ \frac{\sqrt{b} \left (15 a^2 B+20 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 d}+\frac{(-b+i a)^{5/2} (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b (7 a B+4 A b) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{(b+i a)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 2.33047, antiderivative size = 260, 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 = {3607, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{\sqrt{b} \left (15 a^2 B+20 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 d}+\frac{(-b+i a)^{5/2} (-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b (7 a B+4 A b) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{(b+i a)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3607
Rule 3647
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx &=\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{\sqrt{a+b \tan (c+d x)} \left (\frac{1}{2} a (4 a A-b B)+2 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (4 A b+7 a B) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{1}{2} \int \frac{\frac{1}{4} a \left (8 a^2 A-4 A b^2-9 a b B\right )+2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac{1}{4} b \left (20 a A b+15 a^2 B-8 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{4} a \left (8 a^2 A-4 A b^2-9 a b B\right )+2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x+\frac{1}{4} b \left (20 a A b+15 a^2 B-8 b^2 B\right ) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{b \left (20 a A b+15 a^2 B-8 b^2 B\right )}{4 \sqrt{x} \sqrt{a+b x}}+\frac{2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x\right )}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{a^3 A-3 a A b^2-3 a^2 b B+b^3 B+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 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 \left (20 a A b+15 a^2 B-8 b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{3 a^2 A b-A b^3+a^3 B-3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (b \left (20 a A b+15 a^2 B-8 b^2 B\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\left ((a-i b)^3 (i A+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 (b \left (20 a A b+15 a^2 B-8 b^2 B\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{\left (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\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{\sqrt{b} \left (20 a A b+15 a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{4 d}+\frac{b (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac{\left ((a-i b)^3 (i A+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 (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\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} (i A-B) \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 (20 a A b+15 a^2 B-8 b^2 B\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} (i A+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 (4 A b+7 a B) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{4 d}+\frac{b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\\ \end{align*}
Mathematica [A] time = 2.59306, size = 291, normalized size = 1.12 \[ \frac{\frac{\sqrt{a} \sqrt{b} \left (15 a^2 B+20 a A b-8 b^2 B\right ) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a+b \tan (c+d x)}}+b (7 a B+4 A b) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}+4 (-1)^{3/4} (-a-i b)^{5/2} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-4 \sqrt [4]{-1} (a-i b)^{5/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.935, size = 2652267, normalized size = 10201. \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sqrt{\tan \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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]
[Out]
________________________________________________________________________________________
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]
[Out]
________________________________________________________________________________________
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]
[Out]