Optimal. Leaf size=247 \[ \frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-5 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.927366, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3609, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-5 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{\tan (c+d x)}}{3 a^2 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}-\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3609
Rule 3649
Rule 3616
Rule 3615
Rule 93
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \frac{A+B \tan (c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx &=\frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 \int \frac{\frac{1}{2} \left (3 a^2 A+2 A b^2+a b B\right )-\frac{3}{2} a (A b-a B) \tan (c+d x)+b (A b-a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 a \left (a^2+b^2\right )}\\ &=\frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} a^2 \left (a^2 A-A b^2+2 a b B\right )-\frac{3}{4} a^2 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{3 a^2 \left (a^2+b^2\right )^2}\\ &=\frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(A-i B) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}+\frac{(A+i B) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=\frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}+\frac{(A+i B) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=\frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{(A-i B) \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 )}{(a-i b)^2 d}+\frac{(A+i B) \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 )}{(a+i b)^2 d}\\ &=-\frac{(A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}+\frac{2 b (A b-a B) \sqrt{\tan (c+d x)}}{3 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b+2 A b^3-5 a^3 B+a b^2 B\right ) \sqrt{\tan (c+d x)}}{3 a^2 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.46293, size = 273, normalized size = 1.11 \[ \frac{\frac{2 b \left (a^2+b^2\right ) (A b-a B) \sqrt{\tan (c+d x)}}{a (a+b \tan (c+d x))^{3/2}}+\frac{2 b \left (8 a^2 A b-5 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{\tan (c+d x)}}{a^2 \sqrt{a+b \tan (c+d x)}}-3 \sqrt [4]{-1} \left (\frac{i (a-i b)^2 (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}+\frac{(a+i b)^2 (B+i A) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{3 d \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 2.143, size = 2975233, normalized size = 12045.5 \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{B \tan \left (d x + c\right ) + A}{{\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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]